When solving many math problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic ones. The principle of successful solution of each of the mentioned tasks is as follows: it is necessary to establish what type of task is being solved, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

Obviously, success or failure in solving a particular problem depends mainly on how correctly the type of the equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, in this case, it is necessary to have the skills to perform identical transformations and calculations.

A different situation occurs with trigonometric equations. It is not difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer.

It is sometimes difficult to determine its type by the appearance of an equation. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.

To solve the trigonometric equation, we must try:

1. bring all the functions included in the equation to "the same angles";
2. bring the equation to "the same functions";
3. factorize the left side of the equation, etc.

Consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution scheme

Step 1. Express the trigonometric function in terms of known components.

Step 2 Find function argument using formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

sin x = a; x \u003d (-1) n arcsin a + πn, n Є Z.

tan x = a; x \u003d arctg a + πn, n Є Z.

ctg x = a; x \u003d arcctg a + πn, n Є Z.

Step 3 Find an unknown variable.

Example.

2 cos(3x – π/4) = -√2.

Decision.

1) cos(3x - π/4) = -√2/2.

2) 3x – π/4 = ±(π – π/4) + 2πn, n Є Z;

3x – π/4 = ±3π/4 + 2πn, n Є Z.

3) 3x = ±3π/4 + π/4 + 2πn, n Є Z;

x = ±3π/12 + π/12 + 2πn/3, n Є Z;

x = ±π/4 + π/12 + 2πn/3, n Є Z.

Answer: ±π/4 + π/12 + 2πn/3, n Є Z.

II. Variable substitution

Solution scheme

Step 1. Bring the equation to an algebraic form with respect to one of the trigonometric functions.

Step 2 Denote the resulting function by the variable t (if necessary, introduce restrictions on t).

Step 3 Write down and solve the resulting algebraic equation.

Step 4 Make a reverse substitution.

Step 5 Solve the simplest trigonometric equation.

Example.

2cos 2 (x/2) - 5sin (x/2) - 5 = 0.

Decision.

1) 2(1 - sin 2 (x/2)) - 5sin (x/2) - 5 = 0;

2sin 2(x/2) + 5sin(x/2) + 3 = 0.

2) Let sin (x/2) = t, where |t| ≤ 1.

3) 2t 2 + 5t + 3 = 0;

t = 1 or e = -3/2 does not satisfy the condition |t| ≤ 1.

4) sin (x/2) = 1.

5) x/2 = π/2 + 2πn, n Є Z;

x = π + 4πn, n Є Z.

Answer: x = π + 4πn, n Є Z.

III. Equation order reduction method

Solution scheme

Step 1. Replace this equation with a linear one using the power reduction formulas:

sin 2 x \u003d 1/2 (1 - cos 2x);

cos 2 x = 1/2 (1 + cos 2x);

tan 2 x = (1 - cos 2x) / (1 + cos 2x).

Step 2 Solve the resulting equation using methods I and II.

Example.

cos2x + cos2x = 5/4.

Decision.

1) cos 2x + 1/2 (1 + cos 2x) = 5/4.

2) cos 2x + 1/2 + 1/2 cos 2x = 5/4;

3/2 cos 2x = 3/4;

2x = ±π/3 + 2πn, n Є Z;

x = ±π/6 + πn, n Є Z.

Answer: x = ±π/6 + πn, n Є Z.

IV. Homogeneous equations

Solution scheme

Step 1. Bring this equation to the form

a) a sin x + b cos x = 0 (homogeneous equation of the first degree)

or to the view

b) a sin 2 x + b sin x cos x + c cos 2 x = 0 (homogeneous equation of the second degree).

Step 2 Divide both sides of the equation by

a) cos x ≠ 0;

b) cos 2 x ≠ 0;

and get the equation for tg x:

a) a tg x + b = 0;

b) a tg 2 x + b arctg x + c = 0.

Step 3 Solve the equation using known methods.

Example.

5sin 2 x + 3sin x cos x - 4 = 0.

Decision.

1) 5sin 2 x + 3sin x cos x – 4(sin 2 x + cos 2 x) = 0;

5sin 2 x + 3sin x cos x – 4sin² x – 4cos 2 x = 0;

sin 2 x + 3sin x cos x - 4cos 2 x \u003d 0 / cos 2 x ≠ 0.

2) tg 2 x + 3tg x - 4 = 0.

3) Let tg x = t, then

t 2 + 3t - 4 = 0;

t = 1 or t = -4, so

tg x = 1 or tg x = -4.

From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.

Answer: x = π/4 + πn, n Є Z; x \u003d -arctg 4 + πk, k Є Z.

V. Method for transforming an equation using trigonometric formulas

Solution scheme

Step 1. Using all kinds of trigonometric formulas, bring this equation to an equation that can be solved by methods I, II, III, IV.

Step 2 Solve the resulting equation using known methods.

Example.

sinx + sin2x + sin3x = 0.

Decision.

1) (sin x + sin 3x) + sin 2x = 0;

2sin 2x cos x + sin 2x = 0.

2) sin 2x (2cos x + 1) = 0;

sin 2x = 0 or 2cos x + 1 = 0;

From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.

We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.

As a result, x \u003d π / 4 + πn / 2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

Answer: x \u003d π / 4 + πn / 2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

The ability and skills to solve trigonometric equations are very important, their development requires considerable effort, both on the part of the student and the teacher.

Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems, as it were, contains many of the knowledge and skills that are acquired when studying the elements of trigonometry.

Trigonometric equations occupy an important place in the process of teaching mathematics and personality development in general.

Do you have any questions? Don't know how to solve trigonometric equations?
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Solution of the simplest trigonometric equations.

The solution of trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this, the trigonometric circle again turns out to be the best helper.

Recall the definitions of cosine and sine.

The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to rotation by a given angle.

The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to rotation by a given angle.

The positive direction of movement along the trigonometric circle is considered to be movement counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1; 0)

We use these definitions to solve the simplest trigonometric equations.

1. Solve the equation

This equation is satisfied by all such values ​​of the angle of rotation , which correspond to the points of the circle, the ordinate of which is equal to .

Let's mark a point with ordinate on the y-axis:

Draw a horizontal line parallel to the x-axis until it intersects with the circle. We will get two points lying on a circle and having an ordinate. These points correspond to rotation angles of and radians:

If we, having left the point corresponding to the angle of rotation per radian, go around a full circle, then we will come to a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this angle of rotation also satisfies our equation. We can make as many "idle" turns as we like, returning to the same point, and all these angle values ​​will satisfy our equation. The number of "idle" revolutions is denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or ) can take on any integer values.

That is, the first series of solutions to the original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, where , . (2)

As you guessed, this series of solutions is based on the point of the circle corresponding to the angle of rotation by .

These two series of solutions can be combined into one entry:

If we take in this entry (that is, even), then we will get the first series of solutions.

If we take in this entry (that is, odd), then we will get the second series of solutions.

2. Now let's solve the equation

Since is the abscissa of the point of the unit circle obtained by turning through the angle , we mark on the axis a point with the abscissa :

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on a circle and having an abscissa. These points correspond to rotation angles of and radians. Recall that when moving clockwise, we get a negative angle of rotation:

We write down two series of solutions:

,

,

(We get to the right point by passing from the main full circle, that is.

Let's combine these two series into one post:

3. Solve the equation

The line of tangents passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is 1):

Connect this point to the origin with a straight line and mark the points of intersection of the line with the unit circle. The points of intersection of the line and the circle correspond to the rotation angles on and :

Since the points corresponding to the rotation angles that satisfy our equation lie radians apart, we can write the solution as follows:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

We mark a point with the abscissa -1 on the line of cotangents:

Connect this point to the origin of the straight line and continue it until it intersects with the circle. This line will intersect the circle at points corresponding to rotation angles of and radians:

Since these points are separated from each other by a distance equal to , then we can write the general solution of this equation as follows:

In the given examples, illustrating the solution of the simplest trigonometric equations, tabular values ​​of trigonometric functions were used.

However, if there is a non-table value on the right side of the equation, then we substitute the value in the general solution of the equation:

SPECIAL SOLUTIONS:

Mark points on the circle whose ordinate is 0:

Mark a single point on the circle, the ordinate of which is equal to 1:

Mark a single point on the circle, the ordinate of which is equal to -1:

Since it is customary to indicate the values ​​​​closest to zero, we write the solution as follows:

Mark the points on the circle, the abscissa of which is 0:

5.
Let's mark a single point on the circle, the abscissa of which is equal to 1:

Mark a single point on the circle, the abscissa of which is equal to -1:

And some more complex examples:

1.

The sine is one if the argument is

The argument of our sine is , so we get:

Divide both sides of the equation by 3:

Answer:

2.

The cosine is zero if the cosine argument is

The argument of our cosine is , so we get:

We express , for this we first move to the right with the opposite sign:

Simplify the right side:

Divide both parts by -2:

Note that the sign before the term does not change, since k can take any integer values.

Answer:

And in conclusion, watch the video tutorial "Selection of roots in a trigonometric equation using a trigonometric circle"

This concludes the conversation about solving the simplest trigonometric equations. Next time we'll talk about how to solve.

Lesson and presentation on the topic: "Solution of the simplest trigonometric equations"

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What will we study:
1. What are trigonometric equations?

3. Two main methods for solving trigonometric equations.
4. Homogeneous trigonometric equations.
5. Examples.

What are trigonometric equations?

Guys, we have already studied the arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

Trigonometric equations - equations in which the variable is contained under the sign of the trigonometric function.

We repeat the form of solving the simplest trigonometric equations:

1) If |а|≤ 1, then the equation cos(x) = a has a solution:

X= ± arccos(a) + 2πk

2) If |а|≤ 1, then the equation sin(x) = a has a solution:

3) If |a| > 1, then the equation sin(x) = a and cos(x) = a have no solutions 4) The equation tg(x)=a has a solution: x=arctg(a)+ πk

5) The equation ctg(x)=a has a solution: x=arcctg(a)+ πk

For all formulas, k is an integer

The simplest trigonometric equations have the form: Т(kx+m)=a, T- any trigonometric function.

Example.

Solve equations: a) sin(3x)= √3/2

Decision:

A) Let's denote 3x=t, then we will rewrite our equation in the form:

The solution to this equation will be: t=((-1)^n)arcsin(√3/2)+ πn.

From the table of values ​​we get: t=((-1)^n)×π/3+ πn.

Let's go back to our variable: 3x =((-1)^n)×π/3+ πn,

Then x= ((-1)^n)×π/9+ πn/3

Answer: x= ((-1)^n)×π/9+ πn/3, where n is an integer. (-1)^n - minus one to the power of n.

More examples of trigonometric equations.

Solve the equations: a) cos(x/5)=1 b)tg(3x- π/3)= √3

Decision:

A) This time we will go directly to the calculation of the roots of the equation right away:

X/5= ± arccos(1) + 2πk. Then x/5= πk => x=5πk

Answer: x=5πk, where k is an integer.

B) We write in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctg(√3)= π/3

3x- π/3= π/3+ πk => 3x=2π/3 + πk => x=2π/9 + πk/3

Answer: x=2π/9 + πk/3, where k is an integer.

Solve equations: cos(4x)= √2/2. And find all the roots on the segment .

Decision:

Let's solve our equation in general form: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

Now let's see what roots fall on our segment. For k For k=0, x= π/16, we are in the given segment .
With k=1, x= π/16+ π/2=9π/16, they hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means we won’t hit for large k either.

Answer: x= π/16, x= 9π/16

Two main solution methods.

We have considered the simplest trigonometric equations, but there are more complex ones. To solve them, the method of introducing a new variable and the factorization method are used. Let's look at examples.

Let's solve the equation:

Decision:
To solve our equation, we use the method of introducing a new variable, denoted: t=tg(x).

As a result of the replacement, we get: t 2 + 2t -1 = 0

Find the roots of the quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we got the simplest trigonometric equation, let's find its roots.

X=arctg(-1) +πk= -π/4+πk; x=arctg(1/3) + πk.

Answer: x= -π/4+πk; x=arctg(1/3) + πk.

An example of solving an equation

Solve equations: 2sin 2 (x) + 3 cos(x) = 0

Decision:

Let's use the identity: sin 2 (x) + cos 2 (x)=1

Our equation becomes: 2-2cos 2 (x) + 3 cos (x) = 0

2 cos 2 (x) - 3 cos(x) -2 = 0

Let's introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation are the roots: t=2 and t=-1/2

Then cos(x)=2 and cos(x)=-1/2.

Because cosine cannot take values ​​greater than one, then cos(x)=2 has no roots.

For cos(x)=-1/2: x= ± arccos(-1/2) + 2πk; x= ±2π/3 + 2πk

Answer: x= ±2π/3 + 2πk

Homogeneous trigonometric equations.

Definition: An equation of the form a sin(x)+b cos(x) is called homogeneous trigonometric equations of the first degree.

Equations of the form

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, we divide it by cos(x): It is impossible to divide by cosine if it is equal to zero, let's make sure that this is not so:
Let cos(x)=0, then asin(x)+0=0 => sin(x)=0, but sine and cosine are not equal to zero at the same time, we got a contradiction, so we can safely divide by zero.

Solve the equation:
Example: cos 2 (x) + sin(x) cos(x) = 0

Decision:

Take out the common factor: cos(x)(c0s(x) + sin (x)) = 0

Then we need to solve two equations:

cos(x)=0 and cos(x)+sin(x)=0

Cos(x)=0 for x= π/2 + πk;

Consider the equation cos(x)+sin(x)=0 Divide our equation by cos(x):

1+tg(x)=0 => tg(x)=-1 => x=arctg(-1) +πk= -π/4+πk

Answer: x= π/2 + πk and x= -π/4+πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, stick to these rules always!

1. See what the coefficient a is equal to, if a \u003d 0 then our equation will take the form cos (x) (bsin (x) + ccos (x)), an example of the solution of which is on the previous slide

2. If a≠0, then you need to divide both parts of the equation by the squared cosine, we get:

We make the change of variable t=tg(x) we get the equation:

Solve Example #:3

Solve the equation:
Decision:

Divide both sides of the equation by cosine square:

We make a change of variable t=tg(x): t 2 + 2 t - 3 = 0

Find the roots of the quadratic equation: t=-3 and t=1

Then: tg(x)=-3 => x=arctg(-3) + πk=-arctg(3) + πk

Tg(x)=1 => x= π/4+ πk

Answer: x=-arctg(3) + πk and x= π/4+ πk

Solve Example #:4

Solve the equation:

Decision:
Let's transform our expression:

We can solve such equations: x= - π/4 + 2πk and x=5π/4 + 2πk

Answer: x= - π/4 + 2πk and x=5π/4 + 2πk

Solve Example #:5

Solve the equation:

Decision:
Let's transform our expression:

We introduce the replacement tg(2x)=t:2 2 - 5t + 2 = 0

The solution to our quadratic equation will be the roots: t=-2 and t=1/2

Then we get: tg(2x)=-2 and tg(2x)=1/2
2x=-arctg(2)+ πk => x=-arctg(2)/2 + πk/2

2x= arctg(1/2) + πk => x=arctg(1/2)/2+ πk/2

Answer: x=-arctg(2)/2 + πk/2 and x=arctg(1/2)/2+ πk/2

Tasks for independent solution.

1) Solve the equation

A) sin(7x)= 1/2 b) cos(3x)= √3/2 c) cos(-x) = -1 d) tg(4x) = √3 e) ctg(0.5x) = -1.7

2) Solve equations: sin(3x)= √3/2. And find all the roots on the segment [π/2; π].

3) Solve the equation: ctg 2 (x) + 2ctg(x) + 1 =0

4) Solve the equation: 3 sin 2 (x) + √3sin (x) cos(x) = 0

5) Solve the equation: 3sin 2 (3x) + 10 sin(3x)cos(3x) + 3 cos 2 (3x) =0

6) Solve the equation: cos 2 (2x) -1 - cos(x) =√3/2 -sin 2 (2x)

The concept of solving trigonometric equations.

• To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving the trigonometric equation ultimately comes down to solving the four basic trigonometric equations.

Solution of basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tan x = a; ctg x = a
• Solving basic trigonometric equations involves looking at the various x positions on the unit circle, as well as using a conversion table (or calculator).
• Example 1. sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π/3. The unit circle gives another answer: 2π/3. Remember: all trigonometric functions are periodic, that is, their values ​​are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. So the answer is written like this:
• x1 = π/3 + 2πn; x2 = 2π/3 + 2πn.
• Example 2 cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π/3. The unit circle gives another answer: -2π/3.
• x1 = 2π/3 + 2π; x2 = -2π/3 + 2π.
• Example 3. tg (x - π/4) = 0.
• Answer: x \u003d π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x \u003d π / 12 + πn.

Transformations used in solving trigonometric equations.

• To transform trigonometric equations, algebraic transformations (factoring, reduction of homogeneous terms, etc.) and trigonometric identities are used.
• Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.

• Finding angles from known values ​​of functions.

  • Before learning how to solve trigonometric equations, you need to learn how to find angles from known values ​​of functions. This can be done using a conversion table or calculator.
  • Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also equal to 0.732.

• Set aside the solution on the unit circle.

  • You can put solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle are the vertices of a regular polygon.
  • Example: The solutions x = π/3 + πn/2 on the unit circle are the vertices of the square.
  • Example: The solutions x = π/4 + πn/3 on the unit circle are the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If the given trigonometric equation contains only one trigonometric function, solve this equation as a basic trigonometric equation. If this equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
    • Method 1
  • Transform this equation into an equation of the form: f(x)*g(x)*h(x) = 0, where f(x), g(x), h(x) are the basic trigonometric equations.
  • Example 6. 2cos x + sin 2x = 0. (0< x < 2π)
  • Decision. Using the double angle formula sin 2x = 2*sin x*cos x, replace sin 2x.
  • 2cos x + 2*sin x*cos x = 2cos x*(sin x + 1) = 0. Now solve two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
  • Example 7 cos x + cos 2x + cos 3x = 0. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x(2cos x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8. sin x - sin 3x \u003d cos 2x. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x*(2sin x + 1) = 0. Now solve two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
    • Method 2
  • Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x/2) = t, etc.).
  • Example 9. 3sin^2 x - 2cos^2 x = 4sin x + 7 (0< x < 2π).
  • Decision. In this equation, replace (cos^2 x) with (1 - sin^2 x) (according to the identity). The transformed equation looks like:
  • 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation looks like: 5t^2 - 4t - 9 = 0. This is a quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
  • Example 10. tg x + 2 tg^2 x = ctg x + 2
  • Decision. Replace tg x with t. Rewrite the original equation as follows: (2t + 1)(t^2 - 1) = 0. Now find t and then find x for t = tg x.