DEFINITION
Trigonometric inequalities are inequalities that contain a variable under the sign of a trigonometric function.
Solving trigonometric inequalities
The solution of trigonometric inequalities often comes down to solving the simplest trigonometric inequalities of the form: \(\ \sin x a \), \(\ \cos x > a \), \(\ \operatorname(tg) x > a \), \(\ \ operatorname(ctg) x > a \), \(\ \sin x \leq a \), \(\ \cos x \leq a \), \(\ \operatorname(tg) x \leq a \), \ (\ \operatorname(ctg) x \leq a \), \(\ \sin x \geq a \), \(\ \cos \geq a \), \(\ \operatorname(tg) x \geq a \ ), \(\ \operatorname(tg) x \geq a \)
The simplest trigonometric inequalities are solved graphically or using a unit trigonometric circle.
By definition, the sine of the angle \(\ \alpha \) is the ordinate of the point \(\ P_(\alpha)(x, y) \) of the unit circle (Fig. 1), and the cosine is the abscissa of this point. This fact is used in solving the simplest trigonometric inequalities with cosine and sine using the unit circle.
Examples of solving trigonometric inequalities
Solve the inequality \(\ \sin x \leq \frac(\sqrt(3))(2) \)
Since \(\ \left|\frac(\sqrt(3))(2)\right| , this inequality has a solution and can be solved in two ways
First way. Let's solve this inequality graphically. To do this, we construct in the same coordinate system a graph of the sine \(\ y=\sin x \) (Fig. 2) and the straight line \(\ y=\frac(\sqrt(3))(2) \)
Let's select the intervals where the sinusoid is located below the graph of the straight line \(\ y=\frac(\sqrt(3))(2) \) . Find the abscissas \(\ x_(1) \) and \(\ x_(2) \) of the intersection points of these graphs: \(\ x_(1)=\pi-\arcsin \frac(\sqrt(3))(2 )=\pi-\frac(\pi)(3)=\frac(2 \pi)(3) x_(2)=\arcsin \frac(\sqrt(3))(2)+2 \pi=\ frac(\pi)(3)+2 \pi=\frac(7 \pi)(3) \)
We got the interval \(\ \left[-\frac(4 \pi)(3) ; \frac(\pi)(3)\right] \) but since the function \(\ y=\sin x \) is periodic and has a period \(\ 2 \pi \) , then the answer is the union of intervals: \(\ \left[\frac(2 \pi)(3)+2 \pi k ; \frac(7 \pi)(3)+ 2 \pi k\right] \), \(\ k \in Z \)
The second way. Construct a unit circle and a line \(\ y=\frac(\sqrt(3))(2) \) , denote their intersection points \(\ P_(x_(1)) \) and \(\ P_(x_(2 )) \) (Fig. 3). The solution to the original inequality will be the set of ordinate points that are less than \(\ \frac(\sqrt(3))(2) \) . Let's find the value of \(\ \boldsymbol(I)_(1) \) and \(\ \boldsymbol(I)_(2) \) by going counterclockwise, \(\ x_(1) Fig. 3
\(\ x_(1)=\pi-\arcsin \frac(\sqrt(3))(2)=\pi-\frac(\pi)(3)=\frac(2 \pi)(3) x_ (2)=\arcsin \frac(\sqrt(3))(2)+2 \pi=\frac(\pi)(3)+2 \pi=\frac(7 \pi)(3) \)
Taking into account the periodicity of the sine function, we finally obtain the intervals \(\ \left[\frac(2 \pi)(3)+2 \pi k ; \frac(7 \pi)(3)+2 \pi\right] \), \(\k\in Z\)
Solve the inequality \(\ \sin x>2 \)
The sine is a bounded function: \(\ |\sin x| \leq 1 \) , and the right side of this inequality is greater than one, so there are no solutions.
Solve the inequality \(\ \cos x>\frac(1)(2) \)
This inequality can be solved in two ways: graphically and using a unit circle. Let's consider each of the methods.
First way. Let's depict in one coordinate system the functions that describe the left and right parts of the inequality, that is, \(\ y=\cos x \) and \(\ y=\frac(1)(2) \) . Let us select the intervals where the graph of the cosine function \(\ y=\cos x \) is located above the graph of the straight line \(\ y=\frac(1)(2) \) (Fig. 4).
Find the abscissas of the points \(\ \boldsymbol(x)_(1) \) and \(\ x_(2) \) - the points of intersection of the graphs of the functions \(\ y=\cos x \) and \(\ y=\frac (1)(2) \) , which are the ends of one of the intervals on which the indicated inequality holds. \(\ x_(1)=-\arccos \frac(1)(2)=-\frac(\pi)(3) \); \(\ x_(1)=\arccos \frac(1)(2)=\frac(\pi)(3) \)
Considering that the cosine is a periodic function, with a period \(\ 2 \pi \) , the answer is the value \(\ x \) from the intervals \(\ \left(-\frac(\pi)(3)+2 \pi k ; \frac(\pi)(3)+2 \pi k\right) \), \(\ k \in Z \)
The second way. Let's construct a unit circle and a straight line \(\ x=\frac(1)(2) \) (since the x-axis corresponds to the cosines on the unit circle). Let \(\ P_(x_(1)) \) and \(\ P_(x_(2)) \) (Fig. 5) be the intersection points of the line and the unit circle. The solution to the original equation will be the set of abscissa points that are less than \(\ \frac(1)(2) \) . Find the value of \(\ x_(1) \) and \(\ 2 \) , making a counterclockwise tour so that \(\ x_(1) Taking into account the periodicity of the cosine, we finally obtain the intervals \(\ \left(-\frac (\pi)(3)+2 \pi k ;\frac(\pi)(3)+2 \pi k\right) \),\(\ k \in Z \)
Solve the inequality \(\ \operatorname(ctg) x \leq-\frac(\sqrt(3))(3) \)
Let's plot graphs of functions \(\ y=\operatorname(ctg) x \), \(\ y=-\frac(\sqrt(3))(3) \) in one coordinate system
Let's select the intervals where the graph of the function \(\ y=\operatorname(ctg) x \) is not higher than the graph of the straight line \(\ y=-\frac(\sqrt(3))(3) \) (Fig. 6) .
Find the abscissa of the point \(\ x_(0) \) , which is the end of one of the intervals on which the inequality \(\ x_(0)=\operatorname(arcctg)\left(-\frac(\sqrt(3))( 3)\right)=\pi-\operatorname(arcctg)\left(\frac(\sqrt(3))(3)\right)=\pi-\frac(\pi)(3)=\frac(2 \pi)(3) \)
The other end of this gap is the point \(\ \pi \) , and the function \(\ y=\operatorname(ctg) x \) is undefined at this point. Thus, one of the solutions to this inequality is the interval \(\ \frac(2 \pi)(3) \leq x
Trigonometric inequalities with complex argument
Trigonometric inequalities with a complex argument can be reduced to the simplest trigonometric inequalities using a substitution. After solving it, the reverse substitution is made and the original unknown is expressed.
Solve the inequality \(\ 2 \cos \left(2 x+100^(\circ)\right) \leq-1 \)
Express the cosine on the right side of this inequality: \(\ \cos \left(2 x+100^(\circ)\right) \leq-\frac(1)(2) \)
We carry out the replacement \(\ t=2 x+100^(\circ) \) , after which this inequality is transformed to the simplest inequality \(\ \cos t \leq-\frac(1)(2) \)
Let's solve it using the unit circle. Let's construct a unit circle and a line \(\ x=-\frac(1)(2) \) . Let us denote \(\ P_(1) \) and \(\ P_(2) \) as the points of intersection of the line and the unit circle (Fig. 7).
The solution to the original inequality will be the set of abscissa points, which are at most \(\ -\frac(1)(2) \). The point \(\ P_(1) \) corresponds to the angle \(\ 120^(\circ) \) , and the point \(\ P_(2) \) . Thus, given the cosine period, we get \(\ 120^(\circ)+360^(\circ) \cdot n \leq t \leq 240^(\circ)+360^(\circ) \cdot n \) , \(\ n \in Z \)
We make the reverse substitution \(\ t=2 x+100^(\circ) 120^(\circ)+360^(\circ) \cdot n \leq 2 x+100^(\circ) \leq 240^(\ circ)+360^(\circ) \cdot n \), \(\ n \in Z \)
We express \(\ \mathbf(x) \), to do this, first subtract \(\ 100^(\circ) 120^(\circ)-100^(\circ)+360^(\circ) \ cdot n \leq 2 x+100^(\circ)-100^(\circ) \leq 240^(\circ)-100^(\circ)+360^(\circ) \cdot n \), \( \n\in Z\); \(\ 20^(\circ)+360^(\circ) \cdot n \leq 2 x \leq 140^(\circ)+360^(\circ) \cdot n \), \(\ n \in Z\)
and then, divide by 2 \(\ \frac(20^(\circ)+360^(\circ) \cdot n)(2) \leq \frac(2 x)(2) \leq \frac(140^ (\circ)+360^(\circ) \cdot n)(2) \), \(\ n \in Z \); \(\ 10^(\circ)+180^(\circ) \cdot n \leq x \leq 70^(\circ)+180^(\circ) \cdot n \), \(\ n \in Z \)
Double trigonometric inequalities
Solve the double trigonometric inequality \(\ \frac(1)(2)
Let us introduce the replacement \(\ t=\frac(x)(2) \) , then the original inequality will take the form \(\ \frac(1)(2)
Let's solve it using the unit circle. Since the ordinate axis corresponds to the sine on the unit circle, we select on it the set of ordinates of which is greater than \(\ x=\frac(1)(2) \) and less than or equal to \(\ \frac(\sqrt(2))(2 ) \) . In Figure 8, these points will be located on the arcs \(\ P_(t_(1)) \), \(\ P_(t_(2)) \) and \(\ P_(t_(3)) \), \( \ P_(t_(4)) \) . Let's find the value \(\ t_(1) \), \(\ t_(2) \), \(\ t_(3) \), \(\ t_(4) \) , making a counterclockwise tour, and \ (\ t_(1) \(\ t_(3)=\pi-\arcsin \frac(\sqrt(2))(2)=\pi-\frac(\pi)(4)=\frac(3 \ pi)(4) \); \(\ t_(4)=\pi-\arcsin \frac(1)(2)=\pi-\frac(\pi)(6)=\frac(5 \pi) (6)\)
Thus, we obtain two intervals, which, taking into account the periodicity of the sine function, can be written as follows \(\ \frac(\pi)(6)+2 \pi k \leq t \frac(\pi)(4)+2 \ pi k \quad \frac(3 \pi)(4)+2 \pi k leq \frac(x)(2) \frac(\pi)(4)+2 \pi k \), \(\ \frac(3 \pi)(4)+2 \pi k Express \(\ \mathbf( x) \), for this we multiply all sides of both inequalities by 2, we get \(\ \frac(\pi)(3)+4 \pi k \leq x
Inequalities are relations of the form a › b, where a and b are expressions containing at least one variable. Inequalities can be strict - ‹, › and non-strict - ≥, ≤.
Trigonometric inequalities are expressions of the form: F(x) › a, F(x) ‹ a, F(x) ≤ a, F(x) ≥ a, in which F(x) is represented by one or more trigonometric functions.
An example of the simplest trigonometric inequality is: sin x ‹ 1/2. It is customary to solve such problems graphically; two methods have been developed for this.
Method 1 - Solving Inequalities by Plotting a Function
To find an interval that satisfies the conditions of the inequality sin x ‹ 1/2, you must do the following:
- On the coordinate axis, construct a sinusoid y = sin x.
- On the same axis, draw a graph of the numerical argument of the inequality, i.e., a straight line passing through the point ½ of the OY ordinate.
- Mark the intersection points of the two graphs.
- Shade the segment that is the solution of the example.
When there are strong signs in an expression, the intersection points are not solutions. Since the smallest positive period of the sinusoid is 2π, we write the answer as follows:
If the signs of the expression are not strict, then the solution interval must be enclosed in square brackets - . The answer to the problem can also be written as another inequality: 
Method 2 - Solving trigonometric inequalities using the unit circle
Similar problems are easily solved with the help of a trigonometric circle. The search algorithm is very simple:
- First, draw a unit circle.
- Then you need to note the value of the arc function of the argument of the right side of the inequality on the arc of the circle.
- It is necessary to draw a straight line passing through the value of the arc function parallel to the x-axis (OX).
- After that, it remains only to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
- Write the answer in the required form.
Let us analyze the solution steps using the inequality sin x › 1/2 as an example. Points α and β are marked on the circle – the values
The points of the arc located above α and β are the interval for solving the given inequality.
If you need to solve an example for cos, then the arc of answers will be located symmetrically to the OX axis, and not OY. You can consider the difference between the solution intervals for sin and cos in the diagrams below in the text.
Graphical solutions for tangent and cotangent inequalities will differ from both sine and cosine. This is due to the properties of functions.
The arctangent and arccotangent are tangents to the trigonometric circle, and the minimum positive period for both functions is π. In order to quickly and correctly use the second method, you need to remember on which axis the values \u200b\u200bof sin, cos, tg and ctg are plotted.
The tangent tangent runs parallel to the OY axis. If we plot the value of arctg a on the unit circle, then the second required point will be located in the diagonal quarter. corners
They are breakpoints for the function, as the graph tends to them but never reaches them.
In the case of the cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at the points π and 2π.
Complex trigonometric inequalities
If the argument of the inequality function is represented not just by a variable, but by an entire expression containing an unknown, then we are talking about a complex inequality. The course and order of its solution are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:
The graphical solution provides for the construction of an ordinary sinusoid y = sin x for arbitrarily chosen values of x. Let's calculate a table with coordinates for the chart's reference points:
The result should be a nice curve.
For ease of finding a solution, we replace the complex function argument
The intersection of two graphs allows you to determine the area of the desired values for which the inequality condition is satisfied.
The found segment is the solution for the variable t:
However, the goal of the task is to find all possible variants of the unknown x:
Solving the double inequality is quite simple, you need to move π / 3 to the extreme parts of the equation and perform the required calculations:
Answer to the task will look like an interval for strict inequality:
Such tasks will require the experience and skill of students in handling trigonometric functions. The more training tasks will be solved in the process of preparation, the easier and faster the student will find the answer to the question of the exam test.
Solution of the simplest trigonometric equations
First, let's recall the formulas for solving the simplest trigonometric equations.
- $sinx=a$
- $cosx=a$
- $tgx=a$
- $ctgx=a$
Solution of the simplest trigonometric inequalities.
To solve the simplest trigonometric inequalities, we first need to solve the corresponding equation, and then, using the trigonometric circle, find a solution to the inequality. Consider the solutions of the simplest trigonometric inequalities by examples.
Example 1
$sinx\ge \frac(1)(2)$
Find a solution to the trigonometric inequality $sinx=\frac(1)(2)$
\ \
Figure 1. Solution of the inequality $sinx\ge \frac(1)(2)$.
Since the inequality has a “greater than or equal” sign, the solution lies on the upper arc of the circle (with respect to the solution of the equation).
Answer: $\left[\frac(\pi )(6)+2\pi n,\frac(5\pi )(6)+2\pi n\right]$.
Example 2
Find a solution to the trigonometric inequality $cosx=\frac(\sqrt(3))(2)$
\ \
Note the solution on the trigonometric circle
Since the inequality has a “less than” sign, the solution lies on the arc of the circle located to the left (with respect to the solution of the equation).
Answer: $\left(\frac(\pi )(6)+2\pi n,\frac(11\pi )(6)+2\pi n\right)$.
Example 3
$tgx\le \frac(\sqrt(3))(3)$
Find a solution to the trigonometric inequality $tgx=\frac(\sqrt(3))(3)$
\ \
Here we also need a domain of definition. As we remember, the tangent function $x\ne \frac(\pi )(2)+\pi n,n\in Z$
Note the solution on the trigonometric circle
Figure 3. Solution of the inequality $tgx\le \frac(\sqrt(3))(3)$.
Since the inequality has a “less than or equal to” sign, the solution lies on the arcs of the circle marked in blue in Figure 3.
Answer: $\ \left(-\frac(\pi )(2)+2\pi n\right.,\left.\frac(\pi )(6)+2\pi n\right]\cup \left (\frac(\pi )(2)+2\pi n,\right.\left.\frac(7\pi )(6)+2\pi n\right]$
Example 4
Find a solution to the trigonometric inequality $ctgx=\sqrt(3)$
\ \
Here we also need a domain of definition. As we remember, the tangent function $x\ne \pi n,n\in Z$
Note the solution on the trigonometric circle
Figure 4. Solution to the inequality $ctgx\le \sqrt(3)$.
Since the inequality has a “greater than” sign, the solution lies on the arcs of the circle marked in blue in Figure 4.
Answer: $\ \left(2\pi n,\frac(\pi )(6)+2\pi n\right)\cup \left(\pi +2\pi n,\frac(7\pi )( 6)+2\pi n\right)$
METHODS FOR SOLVING TRIGONOMETRIC INEQUALITIES
Relevance. Historically, trigonometric equations and inequalities have been given a special place in the school curriculum. We can say that trigonometry is one of the most important sections of the school course and of all mathematical science in general.
Trigonometric equations and inequalities occupy one of the central places in a high school mathematics course, both in terms of the content of the educational material and the methods of educational and cognitive activity that can and should be formed during their study and applied to solving a large number of problems of a theoretical and applied nature. .
The solution of trigonometric equations and inequalities creates the prerequisites for systematizing students' knowledge related to all educational material in trigonometry (for example, the properties of trigonometric functions, methods for transforming trigonometric expressions, etc.) and makes it possible to establish effective connections with the studied material in algebra (equations, equivalence of equations, inequalities, identical transformations of algebraic expressions, etc.).
In other words, the consideration of methods for solving trigonometric equations and inequalities involves a kind of transfer of these skills to a new content.
The significance of the theory and its numerous applications are proof of the relevance of the chosen topic. This, in turn, allows you to determine the goals, objectives and subject of research of the course work.
Purpose of the study: generalize the available types of trigonometric inequalities, basic and special methods for their solution, select a set of tasks for solving trigonometric inequalities by schoolchildren.
Research objectives:
1. Based on the analysis of the available literature on the research topic, systematize the material.
2. Give a set of tasks necessary to consolidate the topic "Trigonometric inequalities."
Object of study are trigonometric inequalities in the school mathematics course.
Subject of study: types of trigonometric inequalities and methods for their solution.
Theoretical significance is to organize the material.
Practical significance: application of theoretical knowledge in solving problems; analysis of the main frequently encountered methods for solving trigonometric inequalities.
Research methods : analysis of scientific literature, synthesis and generalization of the acquired knowledge, analysis of problem solving, search for optimal methods for solving inequalities.
§one. Types of trigonometric inequalities and basic methods for their solution
1.1. The simplest trigonometric inequalities
Two trigonometric expressions connected by a sign or > are called trigonometric inequalities.
To solve a trigonometric inequality means to find a set of values of the unknowns included in the inequality, under which the inequality is satisfied.
The main part of trigonometric inequalities is solved by reducing them to solving the simplest ones:
This may be a method of factorization, change of variable ( 
, 
etc.), where the usual inequality is first solved, and then the inequality of the form 
etc., or other ways.
The simplest inequalities are solved in two ways: using the unit circle or graphically.
Letf(x
is one of the basic trigonometric functions. To solve the inequality 
it suffices to find its solution on one period, i.e. on any segment whose length is equal to the period of the functionf
x
. Then the solution of the original inequality will be all foundx
, as well as those values that differ from those found by any integer number of periods of the function. In this case, it is convenient to use the graphical method.
Let us give an example of an algorithm for solving inequalities 
(
) and 
.
Algorithm for solving the inequality (
).
1. Formulate the definition of the sine of a numberx on the unit circle.
3. On the y-axis, mark a point with the coordinatea .
4. Through this point, draw a line parallel to the OX axis, and mark the points of intersection of it with the circle.
5. Select an arc of a circle, all points of which have an ordinate less thana .
6. Specify the direction of the bypass (counterclockwise) and write down the answer by adding the period of the function to the ends of the interval2πn
,
.
Algorithm for solving the inequality .
1. Formulate the definition of the tangent of a numberx on the unit circle.
2. Draw a unit circle.
3. Draw a line of tangents and mark a point on it with an ordinatea .
4. Connect this point to the origin and mark the point of intersection of the resulting segment with the unit circle.
5. Select an arc of a circle, all points of which have an ordinate on the tangent line that is less thana .
6. Indicate the direction of the traversal and write down the answer, taking into account the scope of the function, adding a periodpn
,
(the number on the left side of the record is always less than the number on the right side).
Graphical interpretation of solutions to the simplest equations and formulas for solving inequalities in a general form are given in the appendix (Appendices 1 and 2).
Example 1
Solve the inequality 
.
Draw a line on the unit circle 
, which intersects the circle at points A and B.
All valuesy
on the interval NM more
, all points of the arc AMB satisfy this inequality. At all angles of rotation, large 
, but smaller 
,
will take on values greater than
(but not more than one).
Fig.1
Thus, the solution of the inequality will be all values in the interval 
, i.e. 
. In order to get all solutions of this inequality, it is enough to add to the ends of this interval 
, where 
, i.e. 
,
.
Note that the values 
and 
are the roots of the equation 
,
those. 
;
.
Answer: 
, .
1.2. Graphic method
In practice, a graphical method for solving trigonometric inequalities is often useful. Consider the essence of the method on the example of the inequality 
:
1. If the argument is complex (different fromX ), then we replace it witht .
2. We build in one coordinate planetoOy
function graphs 
and 
.
3. We find suchtwo adjacent points of intersection of graphs, between whichsinusoidsituatedabove
straight . Find the abscissas of these points.
4. Write a double inequality for the argumentt , considering the cosine period (t will be between the found abscissas).
5. Do a reverse substitution (return to the original argument) and express the valueX from a double inequality, we write the answer as a numerical interval.
Example 2 Solve the inequality: .
When solving inequalities by a graphical method, it is necessary to build graphs of functions as accurately as possible. Let's transform the inequality to the form:
Let us construct graphs of functions in one coordinate system 
and 
(Fig. 2).
Fig.2
Function graphs intersect at a pointBUT
with coordinates 
; 
. In between 
graph points 
below the chart points 
. And when function values are the same. That's why at 
.
Answer: 
.
1.3. Algebraic Method
Quite often, the original trigonometric inequality, by a well-chosen substitution, can be reduced to an algebraic (rational or irrational) inequality. This method involves transforming the inequality, introducing a substitution, or replacing a variable.
Let's consider the application of this method on specific examples.
Example 3
Reduction to the simplest form 
.
(Fig. 3)
Fig.3
, 
.
Answer: 
,
Example 4 Solve the inequality:
ODZ: 
, 
.
Using formulas: 
, 
we write the inequality in the form: 
.
Or, assuming 
after simple transformations we get
,
,
.
Solving the last inequality by the interval method, we obtain:
Fig.4
, respectively 
. Then from Fig. 4 follows 
, where 
.
Fig.5
Answer: 
, 
.
1.4. Spacing Method
The general scheme for solving trigonometric inequalities by the interval method:
Using trigonometric formulas, factorize.
Find breakpoints and zeros of the function, put them on the circle.
Take any pointTo (but not found earlier) and find out the sign of the product. If the product is positive, then put a point outside the unit circle on the ray corresponding to the angle. Otherwise, put the point inside the circle.
If a point occurs an even number of times, we call it a point of even multiplicity; if an odd number of times, we call it a point of odd multiplicity. Draw arcs as follows: start from a pointTo , if the next point is of odd multiplicity, then the arc intersects the circle at this point, but if the point is of even multiplicity, then it does not intersect.
Arcs behind a circle are positive gaps; inside the circle are negative gaps.
Example 5 Solve the inequality
, 
.
Points of the first series: 
.
Points of the second series: 
.
Each point occurs an odd number of times, that is, all points of odd multiplicity.
Find out the sign of the product at 
: . We mark all points on the unit circle (Fig. 6):
Rice. 6
Answer: 
, 
; 
, 
; 
, 
.
Example 6 . Solve the inequality.
Solution:
Let's find the zeros of the expression .
Getaem :
,
;
, 
;
, 
;
, 
;
On the unit circle, series valuesX
1
represented by dots 
. SeriesX
2
gives points 
. A seriesX
3
we get two points 
. Finally, a seriesX
4
will represent points 
. We put all these points on the unit circle, indicating in parentheses next to each of its multiplicity.
Now let the number 
will be equal. We make an estimate by the sign:
So the pointA
should be chosen on the beam forming the angle 
with beamOh,
outside the unit circle. (Note that the auxiliary beamO
A
it doesn't have to be shown in the picture. DotA
selected approximately.)
Now from the pointA
we draw a wavy continuous line sequentially to all the marked points. And at the points our line passes from one region to another: if it was outside the unit circle, then it passes into it. Approaching the point 
, the line returns to the inner region, since the multiplicity of this point is even. Similarly at the point 
(with an even multiplicity) the line has to be rotated to the outer region. So, we drew a certain picture depicted in Fig. 7. It helps to highlight the desired areas on the unit circle. They are marked with a "+".
Fig.7
Final answer:
Note. If the wavy line, after traversing all the points marked on the unit circle, cannot be returned to the pointA , without crossing the circle in an “illegal” place, this means that an error was made in the solution, namely, an odd number of roots were omitted.
Answer: .
§2. A set of tasks for solving trigonometric inequalities
In the process of developing the ability of schoolchildren to solve trigonometric inequalities, 3 stages can also be distinguished.
1. preparatory,
2. formation of skills to solve the simplest trigonometric inequalities;
3. introduction of trigonometric inequalities of other types.
The purpose of the preparatory stage is that it is necessary to form in schoolchildren the ability to use a trigonometric circle or graph to solve inequalities, namely:
Ability to solve simple inequalities of the form ,
, ,
,
using the properties of the sine and cosine functions;
Ability to make double inequalities for arcs of a numerical circle or for arcs of graphs of functions;
Ability to perform various transformations of trigonometric expressions.
It is recommended to implement this stage in the process of systematizing schoolchildren's knowledge about the properties of trigonometric functions. The main means can be tasks offered to students and performed either under the guidance of a teacher or independently, as well as skills gained in solving trigonometric equations.
Here are examples of such tasks:
1
. Mark a point on the unit circle 
, if
.
2.
In what quarter of the coordinate plane is the point , if 
equals: 
3.
Mark points on the trigonometric circle , if:
4. Bring the expression to trigonometric functionsIquarters.
a) 
,
b) 
,
in) 
5. Given the arc MR.M - middleIth quarter,R - middleIIth quarter. Restrict the value of a variablet for: (compose a double inequality) a) arc MP; b) RM arcs.
6. Write a double inequality for the selected sections of the graph:
Rice. one
7.
Solve inequalities 
, 
, 
, 
.
8. Convert expression .
At the second stage of learning to solve trigonometric inequalities, we can offer the following recommendations related to the methodology for organizing students' activities. At the same time, it is necessary to focus on the students' skills to work with a trigonometric circle or graph, which are formed during the solution of the simplest trigonometric equations.
First, it is possible to motivate the expediency of obtaining a general method for solving the simplest trigonometric inequalities by referring, for example, to an inequality of the form 
.
Using the knowledge and skills acquired at the preparatory stage, students will bring the proposed inequality to the form 
, but may find it difficult to find a set of solutions to the resulting inequality, since it is impossible to solve it only using the properties of the sine function. This difficulty can be avoided by referring to the appropriate illustration (solution of the equation graphically or using a unit circle).
Secondly, the teacher should draw students' attention to different ways of completing the task, give an appropriate example of solving the inequality both graphically and using the trigonometric circle.
Consider such options for solving the inequality .
1. Solving the inequality using the unit circle.
In the first lesson on solving trigonometric inequalities, we will offer students a detailed solution algorithm, which in a step-by-step presentation reflects all the basic skills necessary to solve the inequality.
Step 1.Draw a unit circle, mark a point on the y-axis 
and draw a straight line through it parallel to the x-axis. This line will intersect the unit circle at two points. Each of these points depicts numbers whose sine is equal to .
Step 2This straight line divided the circle into two arcs. Let's single out the one on which numbers are displayed that have a sine greater than . Naturally, this arc is located above the drawn straight line.
Rice. 2
Step 3Let's choose one of the ends of the marked arc. Let's write down one of the numbers that is represented by this point of the unit circle 
.
Step 4In order to choose a number corresponding to the second end of the selected arc, we "pass" along this arc from the named end to the other. At the same time, we recall that when moving counterclockwise, the numbers that we will pass increase (when moving in the opposite direction, the numbers would decrease). Let's write down the number that is depicted on the unit circle by the second end of the marked arc 
.
Thus, we see that the inequality satisfy the numbers for which the inequality 
. We solved the inequality for numbers located on the same period of the sine function. Therefore, all solutions of the inequality can be written as 
Students should be asked to carefully consider the figure and figure out why all the solutions to the inequality can be written in the form 
, 
.
Rice. 3
It is necessary to draw the attention of students to the fact that when solving inequalities for the cosine function, we draw a straight line parallel to the y-axis.
Graphical way to solve the inequality.
Building charts 
and 
, given that 
.
Rice. four
Then we write the equation 
and his solution 
, 
, 
, found using formulas 
, 
, 
.
(Givingn
values 0, 1, 2, we find three roots of the composed equation). Values 
are three consecutive abscissas of the intersection points of the graphs and . Obviously, always on the interval 
the inequality 
, and on the interval 
- inequality 
. We are interested in the first case, and then adding to the ends of this interval a number that is a multiple of the sine period, we obtain a solution to the inequality as: 
, .
Rice. 5
Summarize. To solve the inequality , you need to write the corresponding equation and solve it. From the resulting formula find the roots 
and 
, and write the answer of the inequality in the form: , .
Thirdly, the fact about the set of roots of the corresponding trigonometric inequality is very clearly confirmed when solving it graphically.
Rice. 6
It is necessary to demonstrate to students that the coil, which is the solution to the inequality, repeats through the same interval, equal to the period of the trigonometric function. You can also consider a similar illustration for the graph of the sine function.
Fourthly, it is advisable to carry out work on updating students' methods of converting the sum (difference) of trigonometric functions into a product, to draw the attention of schoolchildren to the role of these techniques in solving trigonometric inequalities.
Such work can be organized through the students' independent fulfillment of the tasks proposed by the teacher, among which we highlight the following:
Fifth, students must be required to illustrate the solution of each simple trigonometric inequality using a graph or a trigonometric circle. Be sure to pay attention to its expediency, especially to the use of a circle, since when solving trigonometric inequalities, the corresponding illustration serves as a very convenient means of fixing the set of solutions to a given inequality
Acquaintance of students with methods for solving trigonometric inequalities that are not the simplest, it is advisable to carry out according to the following scheme: referring to a specific trigonometric inequality referring to the corresponding trigonometric equation joint search (teacher - students) for a solution independent transfer of the found technique to other inequalities of the same type.
In order to systematize students' knowledge of trigonometry, we recommend specifically selecting such inequalities, the solution of which requires various transformations that can be implemented in the process of solving it, focusing students' attention on their features.
As such productive inequalities, we can propose, for example, the following:
In conclusion, we give an example of a set of problems for solving trigonometric inequalities.
1. Solve the inequalities:
2. Solve the inequalities: 3. Find all solutions of inequalities: 4. Find all solutions of inequalities:a) 
, satisfying the condition 
;
b) 
, satisfying the condition 
.
5. Find all solutions of inequalities:
a) ;
b) ;
in) 
;
G) 
;
e) 
.
6. Solve the inequalities:
a) ;
b) ;
in) ;
G) 
;
e) ;
e) ;
and) 
.
7. Solve the inequalities:
a) 
;
b) ;
in) ;
G) .
8. Solve the inequalities:
a) ;
b) ;
in) ;
G) 
;
e) 
;
e) ;
and) 
;
h) .
It is advisable to offer tasks 6 and 7 to students studying mathematics at an advanced level, task 8 - to students in classes with in-depth study of mathematics.
§3. Special methods for solving trigonometric inequalities
Special methods for solving trigonometric equations - that is, those methods that can only be used to solve trigonometric equations. These methods are based on the use of the properties of trigonometric functions, as well as on the use of various trigonometric formulas and identities.
3.1. Sector Method
Consider the sector method for solving trigonometric inequalities. Solution of inequalities of the form 
, whereP
(
x
)
andQ
(
x
)
- rational trigonometric functions (sines, cosines, tangents and cotangents enter them rationally), similarly to the solution of rational inequalities. It is convenient to solve rational inequalities by the method of intervals on the real axis. Its analogue in solving rational trigonometric inequalities is the method of sectors in a trigonometric circle, forsinx
andcosx
(
) or a trigonometric semicircle fortgx
andctgx
(
).
In the interval method, each linear factor of the numerator and denominator of the form 
point on the number axis 
, and when passing through this point changes sign. In the sector method, each multiplier of the form 
, where 
- one of the functionssinx
orcosx
and 
, in a trigonometric circle there correspond two angles 
and 
, which divide the circle into two sectors. When passing through and function changes sign.
The following must be remembered:
a) Multipliers of the form 
and 
, where 
, retain sign for all values 
. Such multipliers of the numerator and denominator are discarded, changing (if 
) for each such rejection, the inequality sign is reversed.
b) Multipliers of the form 
and 
are also discarded. Moreover, if these are factors of the denominator, then inequalities of the form are added to the equivalent system of inequalities 
and 
. If these are factors of the numerator, then in the equivalent system of constraints they correspond to the inequalities and in the case of strict initial inequality, and equality 
and 
in the case of a non-strict initial inequality. When dropping the multiplier 
or 
the inequality sign is reversed.
Example 1
Solve inequalities: a) 
, b) 
.
we have a function, b). Solve the inequality We have
3.2. Concentric circle method
This method is analogous to the method of parallel numerical axes in solving systems of rational inequalities.
Consider an example of a system of inequalities.
Example 5
Solve a system of simple trigonometric inequalities 
First, we solve each inequality separately (Figure 5). In the upper right corner of the figure, we will indicate for which argument the trigonometric circle is considered.
Fig.5
Next, we build a system of concentric circles for the argumentX . We draw a circle and shade it according to the solution of the first inequality, then we draw a circle of a larger radius and shade it according to the solution of the second one, then we build a circle for the third inequality and a base circle. We draw rays from the center of the system through the ends of the arcs so that they intersect all circles. We form a solution on the base circle (Figure 6).
Fig.6
Answer:
, 
.
Conclusion
All objectives of the coursework were completed. The theoretical material is systematized: the main types of trigonometric inequalities and the main methods for their solution (graphical, algebraic, method of intervals, sectors and the method of concentric circles) are given. For each method, an example of solving an inequality was given. The theoretical part was followed by the practical part. It contains a set of tasks for solving trigonometric inequalities.
This coursework can be used by students for independent work. Students can check the level of assimilation of this topic, practice in performing tasks of varying complexity.
Having worked through the relevant literature on this issue, obviously, we can conclude that the ability and skills to solve trigonometric inequalities in the school course of algebra and the beginning of analysis are very important, the development of which requires considerable effort on the part of the mathematics teacher.
Therefore, this work will be useful for teachers of mathematics, as it makes it possible to effectively organize the training of students on the topic "Trigonometric inequalities".
The study can be continued by expanding it to the final qualifying work.
List of used literature
Bogomolov, N.V. Collection of problems in mathematics [Text] / N.V. Bogomolov. – M.: Bustard, 2009. – 206 p.
Vygodsky, M.Ya. Handbook of elementary mathematics [Text] / M.Ya. Vygodsky. – M.: Bustard, 2006. – 509 p.
Zhurbenko, L.N. Mathematics in examples and tasks [Text] / L.N. Zhurbenko. – M.: Infra-M, 2009. – 373 p.
Ivanov, O.A. Elementary mathematics for schoolchildren, students and teachers [Text] / O.A. Ivanov. – M.: MTsNMO, 2009. – 384 p.
Karp, A.P. Tasks in algebra and the beginnings of analysis for the organization of the final repetition and certification in the 11th grade [Text] / A.P. Carp. – M.: Enlightenment, 2005. – 79 p.
Kulanin, E.D. 3000 competitive problems in mathematics [Text] / E.D. Kulanin. – M.: Iris-press, 2007. – 624 p.
Leibson, K.L. Collection of practical tasks in mathematics [Text] / K.L. Leibson. – M.: Bustard, 2010. – 182 p.
Elbow, V.V. Problems with parameters and their solution. Trigonometry: equations, inequalities, systems. Grade 10 [Text] / V.V. Elbow. – M.: ARKTI, 2008. – 64 p.
Manova, A.N. Maths. Express tutor to prepare for the exam: account. allowance [Text] / A.N. Manova. - Rostov-on-Don: Phoenix, 2012. - 541 p.
Mordkovich, A.G. Algebra and beginning of mathematical analysis. 10-11 grades. Textbook for students of educational institutions [Text] / A.G. Mordkovich. – M.: Iris-press, 2009. – 201 p.
Novikov, A.I. Trigonometric functions, equations and inequalities [Text] / A.I. Novikov. - M.: FIZMATLIT, 2010. - 260 p.
Oganesyan, V.A. Methods of teaching mathematics in secondary school: General methodology. Proc. allowance for physics students. - mat. fak. ped. in-comrade. [Text] / V.A. Oganesyan. – M.: Enlightenment, 2006. – 368 p.
Olechnik, S.N. Equations and inequalities. Non-standard solution methods [Text] / S.N. Olekhnik. - M .: Publishing House Factorial, 1997. - 219 p.
Sevryukov, P.F. Trigonometric, exponential and logarithmic equations and inequalities [Text] / P.F. Sevryukov. – M.: National Education, 2008. – 352 p.
Sergeev, I.N. USE: 1000 tasks with answers and solutions in mathematics. All tasks of group C [Text] / I.N. Sergeev. – M.: Exam, 2012. – 301 p.
Sobolev, A.B. Elementary mathematics [Text] / A.B. Sobolev. - Yekaterinburg: GOU VPO USTU-UPI, 2005. - 81 p.
Fenko, L.M. The method of intervals in solving inequalities and studying functions [Text] / L.M. Fenko. – M.: Bustard, 2005. – 124 p.
Friedman, L.M. Theoretical foundations of the methodology of teaching mathematics [Text] / L.M. Friedman. - M .: Book house "LIBROKOM", 2009. - 248 p.
Attachment 1
Graphical interpretation of solutions to the simplest inequalities
Rice. one
Rice. 2
Fig.3
Fig.4
Fig.5
Fig.6
Fig.7
Fig.8
Appendix 2
Solutions to the simplest inequalities
In the practical lesson, we will repeat the main types of tasks from the topic "Trigonometry", we will additionally analyze problems of increased complexity and consider examples of solving various trigonometric inequalities and their systems.
This lesson will help you prepare for one of the types of tasks B5, B7, C1 and C3.
Let's start by repeating the main types of tasks that we reviewed in the Trigonometry topic and solve several non-standard tasks.
Task #1. Convert angles to radians and degrees: a) ; b) .
a) Use the formula for converting degrees to radians
Substitute the given value into it.
b) Apply the formula for converting radians to degrees
Let's perform the substitution 
.
Answer. a) ; b) .
Task #2. Calculate: a) ; b) .
a) Since the angle is far beyond the table, we reduce it by subtracting the period of the sine. Because the angle is given in radians, then the period will be considered as .
b) In this case, the situation is similar. Since the angle is specified in degrees, then we will consider the period of the tangent as .
The resulting angle, although less than the period, is greater, which means that it no longer refers to the main, but to the extended part of the table. In order not to train our memory once again by memorizing an extended table of trigofunction values, we subtract the tangent period again:
We took advantage of the oddness of the tangent function.
Answer. a) 1; b) .
Task #3. Calculate 
, if .
We bring the entire expression to tangents by dividing the numerator and denominator of the fraction by . At the same time, we can not be afraid that, because in this case, the value of the tangent would not exist.
Task #4. Simplify the expression.
The specified expressions are converted using cast formulas. It's just that they are unusually written using degrees. The first expression is generally a number. Simplify all trigofunctions in turn:
Because , then the function changes to a cofunction, i.e. to the cotangent, and the angle falls into the second quarter, in which the sign of the original tangent is negative.
For the same reasons as in the previous expression, the function changes to a cofunction, i.e. to the cotangent, and the angle falls into the first quarter, in which the initial tangent has a positive sign.
Substituting everything into a simplified expression:
Task #5. Simplify the expression.
Let's write the tangent of the double angle according to the corresponding formula and simplify the expression:
The last identity is one of the universal replacement formulas for cosine.
Task #6. Calculate .
The main thing is not to make a standard error and not give an answer that the expression is equal to . It is impossible to use the main property of the arc tangent while there is a factor in the form of a two near it. To get rid of it, we write the expression according to the formula for the tangent of a double angle, while we treat it as an ordinary argument.
Now it is already possible to apply the main property of the arc tangent, remember that there are no restrictions on its numerical result.
Task #7. Solve the equation.
When solving a fractional equation that equates to zero, it is always indicated that the numerator is zero and the denominator is not, because you can't divide by zero.
The first equation is a special case of the simplest equation, which is solved using a trigonometric circle. Think about this solution yourself. The second inequality is solved as the simplest equation using the general formula for the roots of the tangent, but only with the sign not equal.
As we can see, one family of roots excludes another exactly the same family of roots that do not satisfy the equation. Those. there are no roots.
Answer. There are no roots.
Task #8. Solve the equation.
Immediately note that you can take out the common factor and do it:
The equation has been reduced to one of the standard forms, when the product of several factors is equal to zero. We already know that in this case either one of them is equal to zero, or the other, or the third. We write this as a set of equations:
The first two equations are special cases of the simplest ones, we have already met with similar equations many times, so we will immediately indicate their solutions. We reduce the third equation to one function using the double angle sine formula.
Let's solve the last equation separately:
This equation has no roots, because the value of the sine cannot go beyond 
.
Thus, only the first two families of roots are the solution, they can be combined into one, which is easy to show on a trigonometric circle:
 |
This is a family of all halves, i.e.
Let's move on to solving trigonometric inequalities. First, let's analyze the approach to solving an example without using general solution formulas, but with the help of a trigonometric circle.
Task #9. Solve the inequality.
Draw an auxiliary line on the trigonometric circle corresponding to the value of the sine equal to , and show the interval of angles that satisfy the inequality.
 |
It is very important to understand exactly how to specify the resulting angle interval, i.e. what is its beginning and what is its end. The beginning of the gap will be the angle corresponding to the point that we will enter at the very beginning of the gap if we move counterclockwise. In our case, this is the point that is on the left, because moving counterclockwise and passing the right point, on the contrary, we exit the required angle interval. The right point will therefore correspond to the end of the gap.
Now we need to understand the values of the beginning and end angles of our gap of solutions to the inequality. A typical mistake is to immediately indicate that the right point corresponds to the angle , the left and give the answer. This is not true! Please note that we have just indicated the interval corresponding to the upper part of the circle, although we are interested in the lower one, in other words, we have mixed up the beginning and end of the interval of solutions we need.
In order for the interval to start at the corner of the right point and end at the corner of the left point, the first specified angle must be less than the second. To do this, we will have to measure the angle of the right point in the negative reference direction, i.e. clockwise and it will be equal to . Then, starting from it in a positive clockwise direction, we will get to the right point after the left point and get the angle value for it. Now the beginning of the interval of angles is less than the end of , and we can write the interval of solutions without taking into account the period:
Considering that such intervals will repeat an infinite number of times after any integer number of rotations, we get the general solution, taking into account the sine period:
We put round brackets because the inequality is strict, and we puncture the points on the circle that correspond to the ends of the interval.
Compare your answer with the formula for the general solution that we gave in the lecture.
Answer. 
.
This method is good for understanding where the formulas for general solutions of the simplest trigonal inequalities come from. In addition, it is useful for those who are too lazy to learn all these cumbersome formulas. However, the method itself is also not easy, choose which approach to the solution is most convenient for you.
To solve trigonometric inequalities, you can also use the function graphs on which the auxiliary line is built, similarly to the method shown using the unit circle. If you are interested, try to understand this approach to the solution yourself. In what follows, we will use general formulas to solve the simplest trigonometric inequalities.
Task #10. Solve the inequality.
We use the general solution formula, taking into account that the inequality is not strict:
We get in our case:
Answer. 
Task #11. Solve the inequality.
We use the general solution formula for the corresponding strict inequality:
Answer. 
.
Task #12. Solve inequalities: a) ; b) .
In these inequalities, one should not rush to use formulas for general solutions or a trigonometric circle, it is enough just to remember the range of values of sine and cosine.
a) Because 
, then the inequality is meaningless. Therefore, there are no solutions.
b) Because similarly, the sine of any argument always satisfies the inequality specified in the condition. Therefore, the inequality is satisfied by all real values of the argument .
Answer. a) there are no solutions; b) .
Task 13. Solve the inequality 
.
