Solving trigonometric cotangent equations. Basic methods for solving trigonometric equations

Requires knowledge of the basic formulas of trigonometry - the sum of the squares of sine and cosine, the expression of tangent through sine and cosine, and others. For those who have forgotten them or do not know them, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations with the right approach, it’s quite an exciting activity, like, for example, solving a Rubik’s cube.

Based on the name itself, it is clear that a trigonometric equation is an equation in which the unknown is under the sign of the trigonometric function.
There are so-called simple trigonometric equations. Here's what they look like: sinx = a, cos x = a, tan x = a. Let's consider how to solve such trigonometric equations, for clarity, we will use the already familiar trigonometric circle.

sinx = a

cos x = a

tan x = a

cot x = a

Any trigonometric equation is solved in two stages: we reduce the equation to its simplest form and then solve it as a simple trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.

1. Variable substitution and substitution method

Solve the equation 2cos 2 (x + /6) – 3sin( /3 – x) +1 = 0

Using the reduction formulas we get:

2cos 2 (x + /6) – 3cos(x + /6) +1 = 0

Replace cos(x + /6) with y to simplify and get the usual quadratic equation:

2y 2 – 3y + 1 + 0

The roots of which are y 1 = 1, y 2 = 1/2

Now let's go in reverse order

We substitute the found values ​​of y and get two answer options:

2. Solving trigonometric equations through factorization

How to solve the equation sin x + cos x = 1?

Let's move everything to the left so that 0 remains on the right:

sin x + cos x – 1 = 0

Let us use the identities discussed above to simplify the equation:

sin x - 2 sin 2 (x/2) = 0

Let's factorize:

2sin(x/2) * cos(x/2) - 2 sin 2 (x/2) = 0

2sin(x/2) * = 0

We get two equations

3. Reduction to a homogeneous equation

An equation is homogeneous with respect to sine and cosine if all its terms are relative to the sine and cosine of the same degree of the same angle. To solve a homogeneous equation, proceed as follows:

a) transfer all its members to the left side;

b) take all common factors out of brackets;

c) equate all factors and brackets to 0;

d) in brackets a homogeneous equation of a lower degree is obtained, which in turn is divided into a sine or cosine of a higher degree;

e) solve the resulting equation for tg.

Solve the equation 3sin 2 x + 4 sin x cos x + 5 cos 2 x = 2

Let's use the formula sin 2 x + cos 2 x = 1 and get rid of the open two on the right:

3sin 2 x + 4 sin x cos x + 5 cos x = 2sin 2 x + 2cos 2 x

sin 2 x + 4 sin x cos x + 3 cos 2 x = 0

Divide by cos x:

tg 2 x + 4 tg x + 3 = 0

Replace tan x with y and get a quadratic equation:

y 2 + 4y +3 = 0, whose roots are y 1 =1, y 2 = 3

From here we find two solutions to the original equation:

x 2 = arctan 3 + k

4. Solving equations through the transition to a half angle

Solve the equation 3sin x – 5cos x = 7

Let's move on to x/2:

6sin(x/2) * cos(x/2) – 5cos 2 (x/2) + 5sin 2 (x/2) = 7sin 2 (x/2) + 7cos 2 (x/2)

Let's move everything to the left:

2sin 2 (x/2) – 6sin(x/2) * cos(x/2) + 12cos 2 (x/2) = 0

Divide by cos(x/2):

tg 2 (x/2) – 3tg(x/2) + 6 = 0

5. Introduction of auxiliary angle

For consideration, let’s take an equation of the form: a sin x + b cos x = c,

where a, b, c are some arbitrary coefficients, and x is an unknown.

Let's divide both sides of the equation by:



Now the coefficients of the equation, according to trigonometric formulas, have the properties sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let us denote them respectively as cos and sin, where - this is the so-called auxiliary angle. Then the equation will take the form:

cos * sin x + sin * cos x = C

or sin(x + ) = C

The solution to this simplest trigonometric equation is

x = (-1) k * arcsin C - + k, where



It should be noted that the notations cos and sin are interchangeable.

Solve the equation sin 3x – cos 3x = 1

The coefficients in this equation are:

a = , b = -1, so divide both sides by = 2

The main methods for solving trigonometric equations are: reducing the equations to the simplest (using trigonometric formulas), introducing new variables, and factoring. Let's look at their use with examples. Pay attention to the format of writing solutions to trigonometric equations.

A necessary condition for successfully solving trigonometric equations is knowledge of trigonometric formulas (topic 13 of work 6).

Examples.

1. Equations reduced to the simplest.

1) Solve the equation

Solution:

Answer:

2) Find the roots of the equation

(sinx + cosx) 2 = 1 – sinxcosx, belonging to the segment.

Solution:

Answer:

2. Equations that reduce to quadratic.

1) Solve the equation 2 sin 2 x – cosx –1 = 0.

Solution: Using the formula sin 2 x = 1 – cos 2 x, we get

Answer:

2) Solve the equation cos 2x = 1 + 4 cosx.

Solution: Using the formula cos 2x = 2 cos 2 x – 1, we get

Answer:

3) Solve the equation tgx – 2ctgx + 1 = 0

Solution:

Answer:

3. Homogeneous equations

1) Solve the equation 2sinx – 3cosx = 0

Solution: Let cosx = 0, then 2sinx = 0 and sinx = 0 – a contradiction with the fact that sin 2 x + cos 2 x = 1. This means cosx ≠ 0 and we can divide the equation by cosx. We get

Answer:

2) Solve the equation 1 + 7 cos 2 x = 3 sin 2x

Solution:

We use the formulas 1 = sin 2 x + cos 2 x and sin 2x = 2 sinxcosx, we get

sin 2 x + cos 2 x + 7cos 2 x = 6sinxcosx
sin 2 x – 6sinxcosx+ 8cos 2 x = 0

Let cosx = 0, then sin 2 x = 0 and sinx = 0 – a contradiction with the fact that sin 2 x + cos 2 x = 1.
This means cosx ≠ 0 and we can divide the equation by cos 2 x . We get

tg 2 x – 6 tgx + 8 = 0
Let us denote tgx = y
y 2 – 6 y + 8 = 0
y 1 = 4; y2 = 2
a) tgx = 4, x= arctan4 + 2 k, k
b) tgx = 2, x= arctan2 + 2 k, k .

Answer: arctg4 + 2 k, arctan2 + 2 k,k

4. Equations of the form a sinx + b cosx = s, s≠ 0.

1) Solve the equation.

Solution:

Answer:

5. Equations solved by factorization.

1) Solve the equation sin2x – sinx = 0.

Root of the equation f (X) = φ ( X) can only serve as the number 0. Let's check this:

cos 0 = 0 + 1 – the equality is true.

The number 0 is the only root of this equation.

Answer: 0.

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An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tan x` or `ctg x`) is called a trigonometric equation, and it is their formulas that we will consider further.

The simplest equations are called `sin x=a, cos x=a, tg x=a, ctg x=a`, where `x` is the angle to be found, `a` is any number. Let us write down the root formulas for each of them.

1. Equation `sin x=a`.

For `|a|>1` it has no solutions.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=(-1)^n arcsin a + \pi n, n \in Z`

2. Equation `cos x=a`

For `|a|>1` - as in the case of sine, it has no solutions among real numbers.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=\pm arccos a + 2\pi n, n \in Z`

Special cases for sine and cosine in graphs.

3. Equation `tg x=a`

Has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arctg a + \pi n, n \in Z`

4. Equation `ctg x=a`

Also has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arcctg a + \pi n, n \in Z`

Formulas for the roots of trigonometric equations in the table

For sine:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

Solving any trigonometric equation consists of two stages:

• with the help of transforming it to the simplest;
• solve the simplest equation obtained using the root formulas and tables written above.

Let's look at the main solution methods using examples.

Algebraic method.

This method involves replacing a variable and substituting it into an equality.

Example. Solve the equation: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`

`2cos^2(x+\frac \pi 6)-3cos(x+\frac \pi 6)+1=0`,

make a replacement: `cos(x+\frac \pi 6)=y`, then `2y^2-3y+1=0`,

we find the roots: `y_1=1, y_2=1/2`, from which two cases follow:

1. `cos(x+\frac \pi 6)=1`, `x+\frac \pi 6=2\pi n`, `x_1=-\frac \pi 6+2\pi n`.

2. `cos(x+\frac \pi 6)=1/2`, `x+\frac \pi 6=\pm arccos 1/2+2\pi n`, `x_2=\pm \frac \pi 3- \frac \pi 6+2\pi n`.

Answer: `x_1=-\frac \pi 6+2\pi n`, `x_2=\pm \frac \pi 3-\frac \pi 6+2\pi n`.

Factorization.

Example. Solve the equation: `sin x+cos x=1`.

Solution. Let's move all the terms of the equality to the left: `sin x+cos x-1=0`. Using , we transform and factorize the left-hand side:

`sin x — 2sin^2 x/2=0`,

`2sin x/2 cos x/2-2sin^2 x/2=0`,

`2sin x/2 (cos x/2-sin x/2)=0`,

1. `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
2. `cos x/2-sin x/2=0`, `tg x/2=1`, `x/2=arctg 1+ \pi n`, `x/2=\pi/4+ \pi n` , `x_2=\pi/2+ 2\pi n`.

Answer: `x_1=2\pi n`, `x_2=\pi/2+ 2\pi n`.

Reduction to a homogeneous equation

First, you need to reduce this trigonometric equation to one of two forms:

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

Then divide both parts by `cos x \ne 0` - for the first case, and by `cos^2 x \ne 0` - for the second. We obtain equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which need to be solved using known methods.

Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.

Solution. Let's write the right side as `1=sin^2 x+cos^2 x`:

`2 sin^2 x+sin x cos x — cos^2 x=` `sin^2 x+cos^2 x`,

`2 sin^2 x+sin x cos x — cos^2 x -` ` sin^2 x — cos^2 x=0`

`sin^2 x+sin x cos x — 2 cos^2 x=0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right sides by `cos^2 x \ne 0`, we get:

`\frac (sin^2 x)(cos^2 x)+\frac(sin x cos x)(cos^2 x) — \frac(2 cos^2 x)(cos^2 x)=0`

`tg^2 x+tg x — 2=0`. Let's introduce the replacement `tg x=t`, resulting in `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:

1. `tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
2. `tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, ` n \in Z`.

Answer. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.

Moving to Half Angle

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Let's apply the double angle formulas, resulting in: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2 +10 cos^2 x/2`

`4 tg^2 x/2 — 11 tg x/2 +6=0`

Applying the algebraic method described above, we obtain:

1. `tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
2. `tg x/2=3/4`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Answer. `x_1=2 arctg 2+2\pi n, n \in Z`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Introduction of auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, divide both sides by `sqrt (a^2+b^2)`:

`\frac a(sqrt (a^2+b^2)) sin x +` `\frac b(sqrt (a^2+b^2)) cos x =` `\frac c(sqrt (a^2) +b^2))`.

The coefficients on the left side have the properties of sine and cosine, namely the sum of their squares is equal to 1 and their modules are not greater than 1. Let us denote them as follows: `\frac a(sqrt (a^2+b^2))=cos \varphi` , ` \frac b(sqrt (a^2+b^2)) =sin \varphi`, `\frac c(sqrt (a^2+b^2))=C`, then:

`cos \varphi sin x + sin \varphi cos x =C`.

Let's take a closer look at the following example:

Example. Solve the equation: `3 sin x+4 cos x=2`.

Solution. Divide both sides of the equality by `sqrt (3^2+4^2)`, we get:

`\frac (3 sin x) (sqrt (3^2+4^2))+` `\frac(4 cos x)(sqrt (3^2+4^2))=` `\frac 2(sqrt (3^2+4^2))`

`3/5 sin x+4/5 cos x=2/5`.

Let's denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, then we take `\varphi=arcsin 4/5` as an auxiliary angle. Then we write our equality in the form:

`cos \varphi sin x+sin \varphi cos x=2/5`

Applying the formula for the sum of angles for the sine, we write our equality in the following form:

`sin (x+\varphi)=2/5`,

`x+\varphi=(-1)^n arcsin 2/5+ \pi n`, `n \in Z`,

`x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Answer. `x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Fractional rational trigonometric equations

These are equalities with fractions whose numerators and denominators contain trigonometric functions.

Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.

Solution. Multiply and divide the right side of the equality by `(1+cos x)`. As a result we get:

`\frac (sin x)(1+cos x)=` `\frac ((1-cos x)(1+cos x))(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (1-cos^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (sin^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)-` `\frac (sin^2 x)(1+cos x)=0`

`\frac (sin x-sin^2 x)(1+cos x)=0`

Considering that the denominator cannot be equal to zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

Let's equate the numerator of the fraction to zero: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Then `sin x=0` or `1-sin x=0`.

1. `sin x=0`, `x=\pi n`, `n \in Z`
2. `1-sin x=0`, `sin x=-1`, `x=\pi /2+2\pi n, n \in Z`.

Given that ` x \ne \pi+2\pi n, n \in Z`, the solutions are `x=2\pi n, n \in Z` and `x=\pi /2+2\pi n` , `n \in Z`.

Answer. `x=2\pi n`, `n \in Z`, `x=\pi /2+2\pi n`, `n \in Z`.

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, and engineering. Studying begins in the 10th grade, there are always tasks for the Unified State Exam, so try to remember all the formulas of trigonometric equations - they will definitely be useful to you!

However, you don’t even need to memorize them, the main thing is to understand the essence and be able to derive it. It's not as difficult as it seems. See for yourself by watching the video.

The relationships between the basic trigonometric functions - sine, cosine, tangent and cotangent - are specified trigonometric formulas. And since there are quite a lot of connections between trigonometric functions, this explains the abundance of trigonometric formulas. Some formulas connect trigonometric functions of the same angle, others - functions of a multiple angle, others - allow you to reduce the degree, fourth - express all functions through the tangent of a half angle, etc.

In this article we will list in order all the basic trigonometric formulas, which are sufficient to solve the vast majority of trigonometry problems. For ease of memorization and use, we will group them by purpose and enter them into tables.

Page navigation.

Basic trigonometric identities

Basic trigonometric identities define the relationship between sine, cosine, tangent and cotangent of one angle. They follow from the definition of sine, cosine, tangent and cotangent, as well as the concept of the unit circle. They allow you to express one trigonometric function in terms of any other.

For a detailed description of these trigonometry formulas, their derivation and examples of application, see the article.

Reduction formulas

Reduction formulas follow from the properties of sine, cosine, tangent and cotangent, that is, they reflect the property of periodicity of trigonometric functions, the property of symmetry, as well as the property of shift by a given angle. These trigonometric formulas allow you to move from working with arbitrary angles to working with angles ranging from zero to 90 degrees.

The rationale for these formulas, a mnemonic rule for memorizing them and examples of their application can be studied in the article.

Addition formulas

Trigonometric addition formulas show how trigonometric functions of the sum or difference of two angles are expressed in terms of trigonometric functions of those angles. These formulas serve as the basis for deriving the following trigonometric formulas.

Formulas for double, triple, etc. angle

Formulas for double, triple, etc. angle (they are also called multiple angle formulas) show how trigonometric functions of double, triple, etc. angles () are expressed in terms of trigonometric functions of a single angle. Their derivation is based on addition formulas.

More detailed information is collected in the article formulas for double, triple, etc. angle

Half angle formulas

Half angle formulas show how trigonometric functions of a half angle are expressed in terms of the cosine of a whole angle. These trigonometric formulas follow from the double angle formulas.

Their conclusion and examples of application can be found in the article.

Degree reduction formulas

Trigonometric formulas for reducing degrees are designed to facilitate the transition from natural powers of trigonometric functions to sines and cosines in the first degree, but multiple angles. In other words, they allow you to reduce the powers of trigonometric functions to the first.

Formulas for the sum and difference of trigonometric functions

The main purpose formulas for the sum and difference of trigonometric functions is to go to the product of functions, which is very useful when simplifying trigonometric expressions. These formulas are also widely used in solving trigonometric equations, as they allow you to factor the sum and difference of sines and cosines.

Formulas for the product of sines, cosines and sine by cosine

The transition from the product of trigonometric functions to a sum or difference is carried out using the formulas for the product of sines, cosines and sine by cosine.

Universal trigonometric substitution

We complete our review of the basic formulas of trigonometry with formulas expressing trigonometric functions in terms of the tangent of a half angle. This replacement was called universal trigonometric substitution. Its convenience lies in the fact that all trigonometric functions are expressed rationally in terms of the tangent of a half angle without roots.

Bibliography.

• Algebra: Textbook for 9th grade. avg. school/Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova; Ed. S. A. Telyakovsky. - M.: Education, 1990. - 272 pp.: ill. - ISBN 5-09-002727-7
• Bashmakov M. I. Algebra and the beginnings of analysis: Textbook. for 10-11 grades. avg. school - 3rd ed. - M.: Education, 1993. - 351 p.: ill. - ISBN 5-09-004617-4.
• Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
• Gusev V. A., Mordkovich A. G. Mathematics (a manual for those entering technical schools): Proc. allowance.- M.; Higher school, 1984.-351 p., ill.

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