Formulas for writing general solutions of trigonometric equations. Solving trigonometric equations

Chapter 15

15.6. Solving more complex trigonometric equations

In the previous paragraphs 3-5, solutions of the simplest trigonometric equations , , and are given. More complex trigonometric equations containing several trigonometric functions of the same or different arguments are reduced to them through identical transformations or by solving an auxiliary algebraic equation.

The general technique for solving such equations is to replace all the trigonometric functions included in the equation through one function based on formulas connecting these functions. When solving an equation, we strive to make such transformations that lead to equations equivalent to the given one. Otherwise, you need to check the obtained roots.

Losing roots is a common blunder. Other such errors are inaccurate knowledge of the formulas for solving the simplest equations, as well as the inability to correctly find the desired value of the arc function.

Consider examples.

Solve the equation.

Example 2. (example for reduction to one argument).

Solve the equation.

Solution:
It is advisable to move on to the argument. The product is reminiscent of the formula for the sine of a double argument: .
Substituting into the equation, we get: .
On the left side, once again apply the formula for the sine of a double argument, but first multiply both sides of the equation by.
; ; .
We have obtained the simplest type equation and equate the whole argument to the solution of the simplest equation:
, where .

Solve the equation.

Solution:
According to one of the formulas for lowering the degree, we obtain .

After substitution into the equation, we have

Solve the equation.

Solution:
Moving to the right side, we get , which is equal to:
; ; .
Here we had to go by raising the degree of the equation, but we got the opportunity to apply a good solution - to move all the terms into one part and decompose the resulting expression into factors:
.
Equating each factor separately to zero, we obtain a set of equations,

which, as a rule, is equivalent to the given equation (an exception to this rule is considered in the following example).
We solve the equation, we have
, And .
We solve the equation or , we have , and .

Solve the equation.

Including an extraneous root in the answer is considered a blunder. To avoid it, you need to make sure that the obtained roots do not nullify any of the functions that are in the denominator of the fraction of the given equation (if there are fractions) and that with these roots none of the functions in the original equation loses its meaning (if they are included). It should be remembered at what values of the argument the function vanishes and the domain of definition of each trigonometric function. By analogy, they speak of the domain of the equation (the domain of acceptable values, or ODZ, of the unknown). The domain of definition of a trigonometric equation is the common part (intersection) of the domains of definition of the left and right parts of this equation. If the resulting root does not belong to the domain of the equation, then it is an outsider and must be discarded.

solve the equation
.

Solution:
Let's move on to one function. If expressed through , then we get an irrational equation, which is undesirable. Let's replace with:
; .
We solve the resulting equation as a quadratic one with respect to .
or .
The equation has no roots.
For the equation we have:
. But they also mean the same odd numbers, so we write the solution in a simpler way: .

solve the equation
.

To obtain a homogeneous equation (all terms of the same degree - the second), we multiply the right side by the expression, which is equal to.
;
.
Since the roots of the equation are not the roots of the original equation (this can be easily verified by substitution), then, in order to pass to one function, we divide both parts of the equation by .

We solve the quadratic equation for .
or .
For the equation we have: .
For the equation, we get .

Solve the equation.

We express through and , we get
. Here it must be different from zero (otherwise the equation loses its meaning), so the domain of the equation is all . Since , then we multiply both sides of the equation by to get rid of fractions.
;
;
.
For the equation we have

Class: 10

"Equations will exist forever."

A. Einstein

Lesson Objectives:

Educational:
- deepening understanding of methods for solving trigonometric equations;
- to form skills to distinguish, correctly select ways to solve trigonometric equations.
Educational:
- education of cognitive interest in the educational process;
- formation of the ability to analyze the task;
- contribute to the improvement of the psychological climate in the classroom.
Educational:
- to promote the development of the skill of self-acquisition of knowledge;
- encourage students to argue their point of view;

Equipment: poster with basic trigonometric formulas, computer, projector, screen.

1 lesson

I. Actualization of basic knowledge

Orally solve the equations:

1) cosx = 1;
2) 2 cosx = 1;
3) cosx = –;
4) sin2x = 0;
5) sinx = -;
6) sinx = ;
7) tgx = ;
8) cos 2 x - sin 2 x \u003d 0

1) x = 2k;
2) x = ± + 2k;
3) x = ± + 2k;
4) x = k;
5) x \u003d (-1) + k;
6) x \u003d (-1) + 2k;
7) x = + k;
8) x = + k; to Z.

II. Learning new material

- Today we will consider more complex trigonometric equations. Consider 10 ways to solve them. Then there will be two lessons to consolidate, and the next lesson will be a test. At the stand "To the lesson" tasks are posted, similar to which will be on the test work, they must be solved before the test work. (The day before, before the test work, hang out the solutions to these tasks on the stand).

So, we turn to the consideration of methods for solving trigonometric equations. Some of these methods will probably seem difficult to you, while others will be easy, because. you already know some methods of solving equations.

Four students in the class received an individual task: to understand and show you 4 ways to solve trigonometric equations.

(Speaking students prepared slides in advance. The rest of the students in the class write down the main steps in solving equations in a notebook.)

1 student: 1 way. Solving equations by factoring

sin 4x = 3 cos 2x

To solve the equation, we use the formula for the sine of a double angle sin 2 \u003d 2 sin cos
2 sin 2x cos 2x - 3 cos 2x = 0,
cos 2x (2 sin 2x - 3) = 0. The product of these factors is equal to zero if at least one of the factors is equal to zero.

2x = + k, k Z or sin 2x = 1.5 - no solutions, because | sin| 1
x = + k; to Z.
Answer: x = + k, k Z.

2 student. 2 way. Solving equations by converting the sum or difference of trigonometric functions into a product

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

To solve the equation, we use the formula sin–sin = 2 sin cos

cos 3x + 2 sin cos = 0,

cos 3x - 2 sin x cos 3x \u003d 0,

cos 3x (1 - 2 sinx) = 0. The resulting equation is equivalent to the combination of two equations:

The set of solutions of the second equation is completely included in the set of solutions of the first equation. Means

Answer:

3 student. 3 way. Solving equations by converting the product of trigonometric functions into a sum

sin 5x cos 3x = sin 6x cos2x.

To solve the equation, we use the formula

Answer:

4 student. 4 way. Solving Equations Reducing to Quadratic Equations

3 sin x - 2 cos 2 x \u003d 0,
3 sin x - 2 (1 - sin 2 x) \u003d 0,
2 sin 2 x + 3 sin x - 2 = 0,

Let sin x = t, where | t |. We get the quadratic equation 2t 2 + 3t - 2 = 0,

D = 9 + 16 = 25.

Thus . does not satisfy the condition | t |.

So sin x = . That's why .

Answer:

III. Consolidation of what was studied from the textbook by A. N. Kolmogorov

1. No. 164 (a), 167 (a) (quadratic equation)
2. No. 168 (a) (factoring)
3. No. 174 (a) (converting a sum to a product)
4. (convert product to sum)

(At the end of the lesson, show the solution of these equations on the screen for verification)

№ 164 (A)

2 sin 2 x + sin x - 1 = 0.
Let sin x = t, | t | 1. Then
2 t 2 + t - 1 = 0, t = - 1, t= . Where

Answer: - .

№ 167 (A)

3 tg 2 x + 2 tg x - 1 = 0.

Let tg x \u003d 1, then we get the equation 3 t 2 + 2 t - 1 \u003d 0.

Answer:

№ 168 (A)

Answer:

№ 174 (A)

Solve the equation:

Answer:

2 lesson (lesson-lecture)

IV. Learning new material(continuation)

- So, let's continue studying ways to solve trigonometric equations.

5 way. Solution of homogeneous trigonometric equations

Equations of the form a sin x + b cos x = 0, where a and b are some numbers, are called homogeneous equations of the first degree with respect to sin x or cos x.

Consider the equation

sin x – cos x = 0. Divide both sides of the equation by cos x. This can be done, the loss of the root will not occur, because. , If cos x = 0, That sin x = 0. But this contradicts the basic trigonometric identity sin 2 x + cos 2 x = 1.

Get tg x - 1 = 0.

tan x = 1,

Equations of the form a sin 2 x + bcos 2 x + c sin x cos x = 0 , Where a, b, c some numbers are called homogeneous equations of the second degree with respect to sin x or cos x.

Consider the equation

sin 2 x - 3 sin x cos x + 2 cos 2 \u003d 0. We divide both parts of the equation by cos x, and the root will not be lost, because cos x = 0 is not the root of this equation.

tg 2x - 3tgx + 2 = 0.

Let tgx = t. D \u003d 9 - 8 \u003d 1.

Then Hence tg x = 2 or tg x = 1.

As a result x = arctg 2 + , x =

Answer: arctg 2 + ,

Consider another equation: 3 sin 2 x - 3 sin x cos x + 4 cos 2 x = 2.
We transform the right side of the equation in the form 2 = 2 1 = 2 (sin 2 x + cos 2 x). Then we get:
3sin 2 x – 3sin x cos x + 4cos 2 x = 2 (sin 2 x + cos 2 x),
3sin 2 x – 3sin x cos x + 4cos 2 x – 2sin 2 x – 2 cos 2 x = 0,
sin 2 x - 3sin x cos x + 2cos 2 x \u003d 0. (We got the 2nd equation, which we have already analyzed).

Answer: arctg 2 + k,

6 way. Solution of linear trigonometric equations

A linear trigonometric equation is an equation of the form a sin x + b cos x = c, where a, b, c are some numbers.

Consider the equation sin x + cos x= – 1.
Let's rewrite the equation in the form:

Considering that and, we get:

Answer:

7 way. Introduction of an additional argument

Expression a cos x + b sin x can be converted:

(we have already used this transformation when simplifying trigonometric expressions)

We introduce an additional argument - the angle is such that

Then

Consider the equation: 3 sinx + 4 cosx = 1. =

Homework: No. 164 -170 (c, d).

Requires knowledge of the basic formulas of trigonometry - the sum of the squares of the sine and cosine, the expression of the tangent through the 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 put them into 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 a trigonometric function.
There are so-called simple trigonometric equations. Here's what they look like: sinх = a, cos x = a, tg x = a. 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 bring the equation to the simplest form and then solve it as the simplest trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.

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

Let's replace cos(x + /6) with y for simplicity and get the usual quadratic equation:

2y 2 – 3y + 1 + 0

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

Now let's go backwards

We substitute the found values of y and get two answers:

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

We use the above identities to simplify the equation:

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

Let's do the factorization:

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

2sin(x/2) * = 0

We get two equations

Reduction to a homogeneous equation

An equation is homogeneous with respect to sine and cosine if all its terms with respect to sine and cosine are 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) put all common factors out of brackets;

c) equate all factors and brackets to 0;

d) in brackets, a homogeneous equation of a lesser degree is obtained, which, in turn, is divided by a sine or cosine to 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 + 2 cos 2 x

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

Divide by cosx:

tg 2 x + 4 tg x + 3 = 0

We replace tg 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 \u003d arctg 3 + k

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)

Shifting 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

Introduction of an auxiliary angle

For consideration, let's take an equation of the form: a sin x + b cos x \u003d c,

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

Divide both sides of the equation by:

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

cos * sin x + sin * cos x \u003d C

or sin(x + ) = C

The solution to this simple trigonometric equation is

x \u003d (-1) k * arcsin C - + k, where

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

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

In this equation, the coefficients are:

a \u003d, b \u003d -1, so we divide both parts by \u003d 2

Examples:

\(2\sin(⁡x) = \sqrt(3)\)
tg\((3x)=-\) \(\frac(1)(\sqrt(3))\)
\(4\cos^2⁡x+4\sin⁡x-1=0\)
\(\cos⁡4x+3\cos⁡2x=1\)

How to solve trigonometric equations:

Any trigonometric equation should be reduced to one of the following types:

\(\sin⁡t=a\), \(\cos⁡t=a\), tg\(t=a\), ctg\(t=a\)

where \(t\) is an expression with x, \(a\) is a number. Such trigonometric equations are called protozoa. They are easy to solve using () or special formulas:

See infographics on solving simple trigonometric equations here: , and .

Example . Solve the trigonometric equation \(\sin⁡x=-\)\(\frac(1)(2)\).
Solution:

Answer: \(\left[ \begin(gathered)x=-\frac(π)(6)+2πk, \\ x=-\frac(5π)(6)+2πn, \end(gathered)\right.\) \(k,n∈Z\)

What does each symbol mean in the formula for the roots of trigonometric equations, see.

Attention! The equations \(\sin⁡x=a\) and \(\cos⁡x=a\) have no solutions if \(a ϵ (-∞;-1)∪(1;∞)\). Because the sine and cosine for any x is greater than or equal to \(-1\) and less than or equal to \(1\):

\(-1≤\sin x≤1\) \(-1≤\cos⁡x≤1\)

Example . Solve the equation \(\cos⁡x=-1,1\).
Solution: \(-1,1<-1\), а значение косинуса не может быть меньше \(-1\). Значит у уравнения нет решения.
Answer : no solutions.

Example . Solve the trigonometric equation tg\(⁡x=1\).
Solution:

Solve the equation using a number circle. For this:
1) Let's build a circle)
2) Construct the axes \(x\) and \(y\) and the axis of tangents (it passes through the point \((0;1)\) parallel to the axis \(y\)).
3) On the axis of tangents, mark the point \(1\).
4) Connect this point and the origin - a straight line.
5) Note the points of intersection of this line and the number circle.
6)Let's sign the values of these points: \(\frac(π)(4)\) ,\(\frac(5π)(4)\)
7) Write down all the values of these points. Since they are exactly \(π\) apart from each other, all values can be written in one formula:

Answer: \(x=\)\(\frac(π)(4)\) \(+πk\), \(k∈Z\).

Example . Solve the trigonometric equation \(\cos⁡(3x+\frac(π)(4))=0\).
Solution:

Let's use the number circle again.
1) Let's construct a circle, axes \(x\) and \(y\).
2) On the cosine axis (axis \(x\)) mark \(0\).
3) Draw a perpendicular to the cosine axis through this point.
4) Mark the points of intersection of the perpendicular and the circle.
5) Let's sign the values of these points: \(-\) \(\frac(π)(2)\),\(\frac(π)(2)\).
6) Let's write out the entire value of these points and equate them to the cosine (to what is inside the cosine).

\(3x+\)\(\frac(π)(4)\) \(=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\)

\(3x+\)\(\frac(π)(4)\) \(=\)\(\frac(π)(2)\) \(+2πk\) \(3x+\)\(\frac( π)(4)\) \(=-\)\(\frac(π)(2)\) \(+2πk\)

8) As usual, we will express \(x\) in equations.
Remember to treat numbers with \(π\) as well as \(1\), \(2\), \(\frac(1)(4)\), etc. These are the same numbers as all the others. No numerical discrimination!

\(3x=-\)\(\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\) \(3x=-\)\ (\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\)
\(3x=\)\(\frac(π)(4)\) \(+2πk\) \(|:3\) \(3x=-\)\(\frac(3π)(4)\) \(+2πk\) \(|:3\)
\(x=\)\(\frac(π)(12)\) \(+\)\(\frac(2πk)(3)\) \(x=-\)\(\frac(π)( 4)\)\(+\)\(\frac(2πk)(3)\)

Answer: \(x=\)\(\frac(π)(12)\) \(+\)\(\frac(2πk)(3)\) \(x=-\)\(\frac(π)( 4)\) \(+\)\(\frac(2πk)(3)\) , \(k∈Z\).

Reducing trigonometric equations to the simplest ones is a creative task, here you need to use both, and special methods for solving equations:
- Method (the most popular in the exam).
- Method.
- Method of auxiliary arguments.

Consider an example of solving a square-trigonometric equation

Example . Solve the trigonometric equation \(2\cos^2⁡x-5\cos⁡x+2=0\)
Solution:

\(2\cos^2⁡x-5\cos⁡x+2=0\)	Let's make the change \(t=\cos⁡x\).

	Our equation has become typical. You can solve it with .
\(D=25-4 \cdot 2 \cdot 2=25-16=9\)
\(t_1=\)\(\frac(5-3)(4)\) \(=\)\(\frac(1)(2)\) ; \(t_2=\)\(\frac(5+3)(4)\) \(=2\)	We make a replacement.
\(\cos⁡x=\)\(\frac(1)(2)\); \(\cos⁡x=2\)	We solve the first equation using a number circle. The second equation has no solutions since \(\cos⁡x∈[-1;1]\) and cannot be equal to two for any x.
	Let us write down all the numbers lying on at these points.

Answer: \(x=±\)\(\frac(π)(3)\) \(+2πk\), \(k∈Z\).

An example of solving a trigonometric equation with the study of ODZ:

Example (USE) . Solve the trigonometric equation \(=0\)

\(\frac(2\cos^2⁡x-\sin(⁡2x))(ctg x)\)\(=0\)	There is a fraction and there is a cotangent - so you need to write down. Let me remind you that the cotangent is actually a fraction: ctg\(x=\)\(\frac(\cos⁡x)(\sin⁡x)\) Therefore, the DPV for ctg\(x\): \(\sin⁡x≠0\).
ODZ: ctg\(x ≠0\); \(\sin⁡x≠0\) \(x≠±\)\(\frac(π)(2)\) \(+2πk\); \(x≠πn\); \(k,n∈Z\)	Note the "non-solutions" on the number circle.

\(\frac(2\cos^2⁡x-\sin(⁡2x))(ctg x)\)\(=0\)	Let's get rid of the denominator in the equation by multiplying it by ctg\(x\). We can do this because we wrote above that ctg\(x ≠0\).
\(2\cos^2⁡x-\sin⁡(2x)=0\)	Apply the double angle formula for the sine: \(\sin⁡(2x)=2\sin⁡x\cos⁡x\).
\(2\cos^2⁡x-2\sin⁡x\cos⁡x=0\)	If your hands reached out to divide by cosine - pull them back! You can divide by an expression with a variable if it is definitely not equal to zero (for example, such: \(x^2+1,5^x\)). Instead, we take \(\cos⁡x\) out of brackets.
\(\cos⁡x (2\cos⁡x-2\sin⁡x)=0\)	Let's split the equation into two.
\(\cos⁡x=0\); \(2\cos⁡x-2\sin⁡x=0\)	We solve the first equation with using a number circle. Divide the second equation by \(2\) and move \(\sin⁡x\) to the right side.

\(x=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\). \(\cos⁡x=\sin⁡x\)	The roots that turned out are not included in the ODZ. Therefore, we will not write them down in response. The second equation is typical. Divide it by \(\sin⁡x\) (\(\sin⁡x=0\) cannot be a solution to the equation because in this case \(\cos⁡x=1\) or \(\cos⁡ x=-1\)).
	Again we use a circle.
\(x=\)\(\frac(π)(4)\) \(+πn\), \(n∈Z\)	These roots are not excluded by the ODZ, so they can be written as a response.

Answer: \(x=\)\(\frac(π)(4)\) \(+πn\), \(n∈Z\).

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

Solution:

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

Solution:

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 .

Solution:

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:

Solution:
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

Solution:

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

Solution:

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: