The video course "Get an A" includes all the topics you need to successful delivery USE in mathematics for 60-65 points. Completely all tasks 1-13 profile exam mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.

All the necessary theory. Quick Ways solutions, traps and USE secrets. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.

The course contains 5 big topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of exam tasks. Text tasks and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Tricky solutions, useful cheat sheets, development spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solution challenging tasks 2 parts of the exam.

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 square inequalities, fractional equations and equations that reduce to quadratic. Principle successful solution 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 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, it is necessary to have the skills to perform identical transformations and computing.

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.

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

To solve trigonometric equation you should 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 trigonometric function through 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.

Solution.

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 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.

Solution.

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 given equation linear, using the reduction formulas for this:

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.

Solution.

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 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.

Solution.

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 sorts trigonometric formulas, bring this equation to the equation solved by methods I, II, III, IV.

Step 2 Solve the resulting equation using known methods.

Example.

sinx + sin2x + sin3x = 0.

Solution.

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

Trigonometric equations take 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?
To get help from a tutor -.
The first lesson is free!

blog.site, with full or partial copying of the material, a link to the source is required.

Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information at any time when you contact us.

The following are some examples of the types of personal information we may collect and how we may use such information.

What personal information we collect:

• When you submit an application on the site, we may collect various information, including your name, phone number, address Email etc.

How we use your personal information:

• The personal information we collect allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send you important notices and communications.
• We may also use personal information for internal purposes such as auditing, data analysis and various studies to improve the services we provide and to provide you with recommendations regarding our services.
• If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.

Disclosure to third parties

We do not disclose information received from you to third parties.

Exceptions:

• If necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from government agencies on the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public important occasions.
• In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.

Maintaining your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.

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 the problem being solved belongs to, 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.

Solution.

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.

Solution.

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.

Solution.

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.

Solution.

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.

Solution.

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 connected with the solution of trigonometric equations. The process of solving such problems, as it were, contains many 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?
To get the help of a tutor - register.
The first lesson is free!

site, with full or partial copying of the material, a link to the source is required.

More complex trigonometric equations

Equations

sin x = a,
cos x = a,
tg x = a,
ctg x = a

are the simplest trigonometric equations. In this paragraph on concrete examples we will consider more complex trigonometric equations. Their solution, as a rule, is reduced to solving the simplest trigonometric equations.

Example 1 . solve the equation

sin 2 X= cos X sin 2 x.

Transferring all the terms of this equation to the left side and decomposing the resulting expression into factors, we obtain:

sin 2 X(1 - cos X) = 0.

The product of two expressions is equal to zero if and only if at least one of the factors zero, and the other accepts any numerical value, as long as it is defined.

If a sin 2 X = 0 , then 2 X=n π ; X = π / 2n.

If 1 - cos X = 0 , then cos X = 1; X = 2kπ .

So, we got two groups of roots: X = π / 2n; X = 2kπ . The second group of roots is obviously contained in the first, since for n = 4k the expression X = π / 2n becomes
X = 2kπ .

Therefore, the answer can be written in one formula: X = π / 2n, where n-any integer.

Note that this equation could not be solved by reducing by sin 2 x. Indeed, after the reduction, we would get 1 - cos x = 0, whence X= 2k π . Thus, we would lose some roots, for example π / 2 , π , 3π / 2 .

EXAMPLE 2. solve the equation

A fraction is zero only if its numerator is zero.
That's why sin 2 X = 0 , whence 2 X=n π ; X = π / 2n.

From these values X should be discarded as extraneous those values ​​for which sinX vanishes (fractions with zero denominators are meaningless: division by zero is not defined). These values ​​are numbers that are multiples of π . In the formula
X = π / 2n they are obtained for even n. Therefore, the roots of this equation will be the numbers

X = π / 2 (2k + 1),

where k is any integer.

Example 3 . solve the equation

2 sin 2 X+ 7 cos x - 5 = 0.

Express sin 2 X through cosx : sin 2 X = 1 - cos 2x . Then this equation can be rewritten as

2 (1 - cos 2 x) + 7 cos x - 5 = 0 , or

2cos 2 x- 7cos x + 3 = 0.

denoting cosx through at, we come to quadratic equation

2y 2 - 7y + 3 = 0,

whose roots are the numbers 1 / 2 and 3. Hence, either cos x= 1 / 2 or cos X= 3. However, the latter is impossible, since the cosine of any angle in absolute value does not exceed 1.

It remains to be recognized that cos x = 1 / 2 , where

x = ± 60° + 360° n.

Example 4 . solve the equation

2 sin X+ 3cos x = 6.

Because sin x and cos x do not exceed 1 in absolute value, then the expression
2 sin X+ 3cos x cannot take on values ​​greater than 5 . Therefore, this equation has no roots.

Example 5 . solve the equation

sin X+ cos x = 1

By squaring both sides of this equation, we get:

sin 2 X+ 2 sin x cos x+ cos2 x = 1,

but sin 2 X + cos 2 x = 1 . That's why 2 sin x cos x = 0 . If a sin x = 0 , then X = nπ ; if
cos x , then X = π / 2 + kπ . These two groups of solutions can be written in one formula:

X = π / 2n

Since we squared both parts of this equation, it is possible that among the roots we obtained there are extraneous ones. That is why in this example, unlike all the previous ones, it is necessary to make a check. All values

X = π / 2n can be divided into 4 groups

	1) X = 2kπ .	(n=4k)
	2) X = π / 2 + 2kπ .	(n=4k+1)
	3) X = π + 2kπ .	(n=4k+2)
	4) X = 3π / 2 + 2kπ .	(n=4k+3)

At X = 2kπ sin x+ cos x= 0 + 1 = 1. Therefore, X = 2kπ are the roots of this equation.

At X = π / 2 + 2kπ. sin x+ cos x= 1 + 0 = 1 X = π / 2 + 2kπ are also the roots of this equation.

At X = π + 2kπ sin x+ cos x= 0 - 1 = - 1. Therefore, the values X = π + 2kπ are not roots of this equation. Similarly, it is shown that X = 3π / 2 + 2kπ. are not roots.

Thus, this equation has the following roots: X = 2kπ and X = π / 2 + 2mπ., where k and m- any whole numbers.