The roots of the TGX A equation. Trigonometric equations. Factorization

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\u003e\u003e Arcthangence and Arkkothangenes. Solution of equations TGX \u003d A, CTGX \u003d A

§ 19. Arctanens and Arkotanens. Solution of equations TGX \u003d A, CTGX \u003d A

In Example 2 §16, we could not solve three equations:

Two of them we have already decided - the first in § 17 and the second in § 18, for this we had to introduce the concepts arkkosinus And Arksinus. Consider the third equation x \u003d 2.
The graphs of functions y \u003d Tg x and y \u003d 2 have infinitely many common points, the abscissa of all these points have the form - the abscissa of the intersection points of the direct y \u003d 2 with the main branch of the tangensoid (Fig. 90). For the number X1 mathematics, the designation of AGSTG 2 was invented ("Arcthangence of the two"). Then all the roots of the equation x \u003d 2 can be described by the formula X \u003d AGSTG 2 + PC.
What is AGSTG 2? This is a number tangent which is 2 and which belongs to the interval
Consider now the TG X \u003d -2 equation.
Functions graphics have infinitely many common points, the abscissa of all these points are viewed The abscissa point of the intersection of the straight y \u003d -2 with the main branch of TangentSoids. For the number x 2 mathematics, the designation of AGSTG (-2) was invented. Then all the roots of the equation x \u003d -2 can be described by the formula

What is AGSTG (-2)? This number, whose tangent is -2 and which belongs to the interval. Please note (see Fig. 90): x 2 \u003d -x 2. This means that AGSTG (-2) \u003d - AGSTG 2.
We formulate the definition of Arctgennes in general.

Definition 1. AGSTG A (ARCTANGENS A) is such a number from the interval whose tangent is a. So,

Now we are able to make a general conclusion about the decision equations x \u003d A: equation x \u003d a has solutions

Above, we noted that AGSTG (-2) \u003d -AGSTG 2. In general, for any value, but the formula is valid

Example 1. Calculate:

Example 2. Solve equations:

A) make a formula of solutions:

We cannot calculate the value of Arctangent in this case, therefore, the recording of solutions of the equation will be left in the resulting form.
Answer:
Example 3. Solve inequalities:
The inequality of the species can be solved graphically by following the following plans.
1) construct tangentsoid y \u003d tg x and straight y \u003d a;
2) allocate for the main branch of the tangeysoids of the axis of the X axis, on which the specified inequality is performed;
3) Given the frequency of the function y \u003d tg x, write down the answer in general.
Apply this plan to solve the specified inequalities.

: a) We construct graphs of functions y \u003d tgx and y \u003d 1. On the main branch of the tangensoids, they intersect at the point

We highlight the gap of the x axis, on which the main branch of the tangensoid is located below the direct y \u003d 1, is the interval
Considering the frequency of the function y \u003d TGX, we conclude that the specified inequality is performed at any interval of the form:

Combining all such intervals and represents a general solution of the specified inequality.
The answer can be recorded differently:

b) We construct graphs of functions y \u003d tg x and y \u003d -2. On the main branch of the TangentSoids (Fig. 92), they intersect at the point X \u003d AGSTG (-2).

We highlight the gap of the X axis on which the main branch of tangensoids

Consider the equation with TG X \u003d A, where a\u003e 0. The graphs of the functions y \u003d CTG x and y \u003d and have infinitely many common points, the abscissas of all these points are of the form: x \u003d x 1 + PC, where x 1 \u003d agssort A - abscissa of the intersection points direct y \u003d a with the main branch of tangentsoids (rice . 93). It means that the agssr a is the number whose catangent is equal to and which belongs to the interval (0, P); At this interval, the main branch of the graph of the function y \u003d stg x is being built.

In fig. 93 shows the graphic illustration of solving the C1TG \u003d -A equation. Graphs of functions y \u003d stg x and y \u003d -a have infinitely many common points, the abscissas of all these points are of the form x \u003d x 2 + pc, where x 2 \u003d agssort (- a) - the abscissa of the intersection points direct y \u003d -A TangentSoid branch. It means that the agssort (-a) is the number whose catangent is -a and which belongs to the interval (O, P); At this interval, the main branch of the graph of the function y \u003d stg x is being built.

Definition 2.aSTG A (Arkkothangence A) is such a number from the interval (0, P), whose catangent is equal to a.
So,

Now we are able to make a general conclusion about the solution of the CTG equation x \u003d A: the CTG equation X \u003d A has solutions:

Please note (see Fig. 93): X 2 \u003d P-X 1. It means that

Example 4. Calculate:

A) put it

The CTG equation X \u003d A is almost always possible to convert an exception to the CTG equation x \u003d 0. But in this case, using the fact that you can go to
cOs x \u003d 0 equation. Thus, the equation of the species X \u003d and independent interest does not represent.

A.G. Mordkovich Algebra Grade 10

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Previously, according to the program, students received an idea of \u200b\u200bsolving trigonometric equations, familiarized with the concepts of Arkkosinus and Arksinus, examples of solutions of the COS T \u003d A and SIN equations. In this video language, consider the solution of equations TG x \u003d a and Ctg x \u003d a.

At the beginning of the study of this topic, consider the equations TG x \u003d 3 and Tg x \u003d - 3. If the TG x \u003d 3 equation is solved using the graphics, we will see that the intersection of the functions Y \u003d TG X and Y \u003d 3 has infinite set solutions, where x \u003d x 1 + πk. The value x 1 is the coordinate x point of intersection of the graphs of functions Y \u003d TG X and Y \u003d 3. The author introduces the concept of Arctangent: Arctg 3 is the number, tg of which is 3, and this number belongs to the interval from -π / 2 to π / 2. Using the concept of Arctangent, the solution of the TG x \u003d 3 equation can be written in the form x \u003d arctg 3 + πk.

By analogy, the TG x \u003d - - 3 equation is solved according to the structures of functions y \u003d Tg x and y \u003d - 3 it can be seen that the points of intersection of graphs, and, consequently, the solutions of the equations will be x \u003d x 2 + πk. With the help of Arctangent, the solution can be written as x \u003d arctg (- 3) + πk. In the following figure, see what arctg (- 3) \u003d - arctg 3.

The general definition of Arctangent looks like this: Arctangent A is called such a number from the gap from -π / 2 to π / 2, the tangent of which is equal to a. Then by the solution of the TG x \u003d A equation is x \u003d arctg a + πk.

The author brings an example 1. Find the ARCTG expression solution. We appreciate: the number of X is x, then TG X will be equal to a given number, where X belongs to the cut from -π / 2 to π / 2. As in the examples in previous topics, we use the table of values. On this table, the tangent of this number corresponds to x \u003d π / 3. We write down the solution of the ARCTANHANCENS number equation is π / 3, π / 3 belongs to the interval from -π / 2 to π / 2.

Example 2 - Calculate the arctangent of a negative number. Using the ARCTG equality (- a) \u003d - Arctg A, we introduce the x value. Analogously to Example 2 Write the value of x, which belongs to the segment from -π / 2 to π / 2. On the table of the values \u200b\u200bwe find that x \u003d π / 3, therefore, TG x \u003d - π / 3. The response of the equation will be π / 3.

Consider Example 3. Let the equation TG x \u003d 1. Write that x \u003d arctg 1 + πk. In the table, the value of TG 1 corresponds to x \u003d π / 4, therefore, arctg 1 \u003d π / 4. We substitute this value to the original X formula and write the answer x \u003d π / 4 + πk.

Example 4: Calculate TG X \u003d - 4.1. In this case, x \u003d arctg (- 4.1) + πk. Because It is not possible to find the ARCTG value in this case, the answer will look like x \u003d arctg (- 4.1) + πk.

In Example 5, the solution of the inequality TG X\u003e 1. To solve, construct graphs of functions Y \u003d TG X and Y \u003d 1. As can be seen in the figure, these graphs intersect at points x \u003d π / 4 + πk. Because In this case, TG X\u003e 1, on the chart, select the area of \u200b\u200bthe tangensoid, which is above the graph Y \u003d 1, where X belongs to the interval from π / 4 to π / 2. Answer write as π / 4 + πk< x < π/2 + πk.

Next, consider the ctg x \u003d a equation. The figure shows graphs of functions y \u003d CTG X, Y \u003d A, Y \u003d - A, which have a variety of intersection points. Decisions can be written as x \u003d x 1 + πk, where x 1 \u003d arcctg a and x \u003d x 2 + πk, where x 2 \u003d arcctg (- a). It is noted that x 2 \u003d π - x 1. From this follows the equality of ArcCTG (- a) \u003d π - ArcCTG a. Next is given to the definition of Arkkothangence: Arkkothangent A is called such a number from the gap from 0 to π, whose catangent is equal to a. The solution of the CTG X \u003d A equation is written in the form: X \u003d ArcCTG A + πK.

At the end of the video tutorial, another important output is made - the expression Ctg X \u003d A can be written in the form TG x \u003d 1 / a, provided that A is not zero.

Text decoding:

Consider the solution of the equations TG x \u003d 3 and Tg x \u003d - 3. Solving the first equation graphically, we see that the graphs of the functions y \u003d TG x and y \u003d 3 have infinitely many points of intersection, which will write the abscissions in the form of

x \u003d x 1 + πk, where x 1 is the abscissa of the intersection points of the straight y \u003d 3 with the main branch of TangentSoids (Fig. 1), for which the designation was invented

arctg 3 (Arctgernes Three).

How to understand arctg 3?

This number, whose tangent is 3 and this number belongs to the interval (-;). Then all the roots of the TG equation x \u003d 3 can be written by the formula x \u003d arctg 3 + πk.

Similarly, the solution of the TG equation x \u003d - 3 can be written in the form x \u003d x 2 + πk, where x 2 is the abscissa of the intersection points of the straight y \u003d - 3 with the main branch of the TangentSoids (Fig. 1), for which the ARCTG designation was invented (- 3) (Arctangen minus three). Then all the roots of the equation can be recorded by the formula: x \u003d arctg (-3) + πk. Figure shows that Arctg (- 3) \u003d - Arctg 3.

We formulate the definition of Arctangent. Arctangent A is called such a number from the gap (-;), whose tangent is equal to a.

Often use equality: arctg (-a) \u003d -arctg A, which is valid for anyone a.

Knowing the definition of Arctangent, we will make a general conclusion about solving the equation

tG x \u003d A: The equation TG x \u003d A has the solution x \u003d arctg a + πk.

Consider examples.

Example 1. Complimentary arctg.

Decision. Let arctg \u003d x, then TGX \u003d and Xε (-;). Show the table of values \u200b\u200bconsequently, x \u003d, since TG \u003d and ε (-;).

So, arctg \u003d.

Example 2. Calculate ArctG (-).

Decision. Using the ARCTG equality (- a) \u003d - arctg a, write:

arctg (-) \u003d - arctg. Let - arctg \u003d x, then - tgh \u003d and xε (-;). Consequently, x \u003d, since Tg \u003d and ε (-;). Show table values

So - arctg \u003d - TGX \u003d -.

Example 3. Solve equation TGX \u003d 1.

1. We write down the solutions formula: x \u003d arctg 1 + πk.

2. Find the meaning of Arctangent

since TG \u003d. Show table values

So arctg1 \u003d.

3. Put the value found in the decision formula:

Example 4. Solve equation TGX \u003d - 4.1 (Tangent X is equal to minus four integer one tenth).

Decision. We write the solutions formula: x \u003d arctg (- 4.1) + πk.

We cannot calculate the value of Arctangent, therefore, the solution of the equation will be left in the resulting form.

Example 5. Solve the inequality TGX 1.

Decision. We will decide graphically.

1. Build TangentSoid

y \u003d tgh and straight y \u003d 1 (Fig.2). They intersect at the points of the species x \u003d + πk.

2. We highlight the gap of the X axis, on which the main branch of the tangentsoid is located above the straight line y \u003d 1, since under the condition TGX 1. This is an interval (;).

3. Use the frequency of the function.

Nivestly 2. y \u003d Tg x is a periodic function with the main period π.

Given the frequency of the function y \u003d TGX, we write the answer:

(;). The answer can be written in the form of double inequality:

Let us turn to the CTG x \u003d a equation. Imagine a graphic illustration of solving the equation for a positive and negative A (Fig. 3).

Graphs of functions y \u003d ctg x and y \u003d as well

y \u003d ctg x and y \u003d -a

there are infinitely many common points whose abscissions are:

x \u003d x 1 +, where x 1 is the abscissa point of the intersection of direct y \u003d a with the main branch of tangentsoids and

x 1 \u003d arcstg a;

x \u003d x 2 +, where x 2 is the abscissa point of intersection

y \u003d - and with the main branch of tangentzoids and x 2 \u003d arcstg (- a).

Note that x 2 \u003d π - x 1. So we write important equality:

aRCSTG (-A) \u003d π - ArcStg a.

We formulate the definition: Arkkothangent A is called such a number from the interval (0; π), the catangent of which is equal to a.

The solution of the CTG equation x \u003d a is written in the form: x \u003d arcstg a +.

Note that the CTG equation x \u003d a can be converted to mind

tG x \u003d, for the exception, when a \u003d 0.

To successfully decide trigonometric equations Convenient to use method informationto previously solved tasks. Let's figure it out, what is the essence of this method?

In any proposed task, you need to see the previously resolved task, and then with the help of successive equivalent transformations, try to reduce the task to you simple.

Thus, when solving trigonometric equations, they usually constitute some finite sequence of equivalent equations, the last link of which is the equation with an obvious solution. It is only important to remember that if the skills of solving the simplest trigonometric equations are not formed, the solution of more complex equations will be difficult and ineffective.

In addition, solving trigonometric equations should never be forgotten about the possibility of the existence of several ways to solve.

Example 1. Find the number of roots of the COS X \u003d -1/2 equation on the interval.

Decision:

I Method. I will depict the graphs of the functions y \u003d cos x and y \u003d -1/2 and we find the number of their common points on the interval (Fig. 1).

Since the graphs of functions have two common points on the interval, the equation contains two roots at a given interval.

II way. Via trigonometric circle (Fig. 2) Find out the number of points belonging to the gap in which COS X \u003d -1/2. Figure shows that the equation has two roots.

III way. Taking advantage of the formula of the roots of the trigonometric equation, resolving the COS X \u003d -1/2 equation.

x \u003d ± Arccos (-1/2) + 2πk, k is an integer (K \u200b\u200b€ Z);

x \u003d ± (π - arccos 1/2) + 2πk, k is an integer (k € Z);

x \u003d ± (π - π / 3) + 2πk, k is an integer (k € z);

x \u003d ± 2π / 3 + 2πk, k is an integer (k € Z).

The gap belongs to the roots 2π / 3 and -2π / 3 + 2π, k - an integer. Thus, the equation has two roots at a given interval.

Answer: 2..

In the future, trigonometric equations will be solved by one of the proposed methods, which in many cases does not exclude the application and other ways.

Example 2. Find the number of solutions of the TG equation (x + π / 4) \u003d 1 at the interval [-2π; 2π].

Decision:

Using the formula of the roots of the trigonometric equation, we obtain:

x + π / 4 \u003d arctg 1 + πk, k is an integer (k € z);

x + π / 4 \u003d π / 4 + πk, k is an integer (k € Z);

x \u003d πk, k is an integer (k € Z);

Interval [-2π; 2π] belong to the number -2π; -π; 0; π; 2π. So, the equation has five roots at a given interval.

Answer: 5.

Example 3. Find the number of roots of the COS 2 x + sin x · cos x \u003d 1 equation at the interval [-π; π].

Decision:

Since 1 \u003d sin 2 x + cos 2 x (basic trigonometric identity), the initial equation takes the form:

cos 2 x + sin x · cos x \u003d sin 2 x + cos 2 x;

sin 2 x - sin x · cos x \u003d 0;

sIN X (SIN X - COS X) \u003d 0. The product is zero, which means at least one of the multipliers should be zero, so:

sin X \u003d 0 or SIN X - COS X \u003d 0.

Since the value of the variable in which cos x \u003d 0 is not the roots of the second equation (sinus and cosine of the same number cannot simultaneously be zero), then we split both parts of the second equation on COS X:

sIN X \u003d 0 or SIN X / COS X - 1 \u003d 0.

In the second equation, we use that TG X \u003d SIN X / COS X, then:

sin x \u003d 0 or tg x \u003d 1. With the help of formulas, we have:

x \u003d πk or x \u003d π / 4 + πk, k is an integer (k € Z).

From the first series of roots, the interval [-π; π] belong numbers -π; 0; π. From the second series: (π / 4 - π) and π / 4.

Thus, the five roots of the initial equation belong to the interval [-π; π].

Answer: 5.

Example 4. Find the amount of the roots of the TG 2 X + CTG 2 x + 3tg x + 3 x + 3tg x + 3 ctgx + 4 \u003d 0 on the interval [-π; 1,1π].

Decision:

We rewrite the equation in the following form:

tG 2 X + CTG 2 x + 3 (TG X + CTGX) + 4 \u003d 0 and make a replacement.

Let TG X + CTGX \u003d a. Both parts of equality are erected into a square:

(TG X + CTG x) 2 \u003d a 2. Recall brackets:

tG 2 x + 2tg x · CTGX + CTG 2 x \u003d A 2.

Since TG x · stgx \u003d 1, then TG 2 x + 2 + CTG 2 x \u003d a 2, and therefore

tG 2 X + CTG 2 x \u003d A 2 - 2.

Now the initial equation is:

a 2 - 2 + 3a + 4 \u003d 0;

a 2 + 3A + 2 \u003d 0. With the help of the Viiet theorem, we obtain that a \u003d -1 or a \u003d -2.

We will replace replacement, we have:

tG X + CTGX \u003d -1 or TG X + CTGX \u003d -2. We solve the obtained equations.

tG X + 1 / TGX \u003d -1 or TG X + 1 / TGX \u003d -2.

By the property of two mutually reverse numbers, we determine that the first equation does not have roots, and from the second equation we have:

tG x \u003d -1, i.e. x \u003d -π / 4 + πk, k is an integer (K \u200b\u200b€ Z).

Interval [-π; 1,1π] belong to the roots: -π / 4; -π / 4 + π. Their amount:

-π / 4 + (-π / 4 + π) \u003d -π / 2 + π \u003d π / 2.

Answer: π / 2.

Example 5. Find the average arithmetic roots of the SIN 3X + SIN X \u003d SIN 2X equation in the interval [-π; 0.5π].

Decision:

We use the sin α + sin β \u003d 2Sin formula ((α + β) / 2) · cos ((α - β) / 2), then

sIN 3X + SIN X \u003d 2SIN ((3x + x) / 2) · COS ((3x - x) / 2) \u003d 2Sin 2X · COS X and the equation takes the view

2Sin 2X · COS X \u003d SIN 2X;

2Sin 2X · COS X - SIN 2X \u003d 0. I will summarize the SIN 2x multiplier for brackets

sIN 2X (2COS X - 1) \u003d 0. Resolved the resulting equation:

sin 2x \u003d 0 or 2cos x - 1 \u003d 0;

sin 2x \u003d 0 or cos x \u003d 1/2;

2x \u003d πk or x \u003d ± π / 3 + 2πk, k is an integer (K \u200b\u200b€ Z).

So we have roots

x \u003d πk / 2, x \u003d π / 3 + 2πk, x \u003d -π / 3 + 2πk, k is an integer (K \u200b\u200b€ Z).

Interval [-π; 0,5π] belong to the roots -π; -π / 2; 0; π / 2 (from the first series of roots); π / 3 (from the second series); -π / 3 (from the third series). Their arithmetic average is:

(-π - π / 2 + 0 + π / 2 + π / 3 - π / 3) / 6 \u003d -π / 6.

Answer: -π / 6.

Example 6. Find the number of roots of the SIN X + COS x \u003d 0 equation on the interval [-1.25π; 2π].

Decision:

This equation is a homogeneous equation of the first degree. We split both parts on COSX (the value of the variable in which cos x \u003d 0 is not the roots of this equation, since the sinus and cosine of the same number cannot simultaneously be zero). The initial equation has the form:

x \u003d -π / 4 + πk, k is an integer (K \u200b\u200b€ Z).

Interval [-1.25π; 2π] belong to the roots -π / 4; (-π / 4 + π); and (-π / 4 + 2π).

Thus, the three root of the equation belongs to the specified gap.

Answer: 3.

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Equality containing an unknown trigonometric function (`sin x, cos x, tg x` or` CTG x`) is called the trigonometric equation, we will consider their formulas further.

The simplest are called the equations `sin x \u003d a, cos x \u003d a, tg x \u003d a, ctg x \u003d a`, where` x` is the angle to find,` a` - any number. We write for each of them the formula roots.

1. Equation `SIN X \u003d A`.

With `| A |\u003e 1` Do not have solutions.

With `| A | \\ LEQ 1` has an infinite number of solutions.

Formula roots: `x \u003d (- 1) ^ n Arcsin A + \\ PI N, N \\ in z`

2. Equation `COS X \u003d A`

With `| a |\u003e 1` - as in the case of sinus, solutions among valid numbers has no.

With `| A | \\ LEQ 1` has infinite set solutions.

Formula roots: `x \u003d \\ pm Arccos A + 2 \\ PI N, N \\ in z`

Private cases for sinus and cosine in charts.

3. Equation `TG X \u003d A`

It has an infinite set of solutions for any values \u200b\u200bof `a`.

Formula of the roots: `x \u003d arctg a + \\ pi n, n \\ in z`

4. Equation `CTG X \u003d A`

It also has infinite set solutions for any values \u200b\u200bof `a`.

Formula roots: `X \u003d ArcCTG A + \\ PI N, N \\ In Z`

The formulas of the roots of trigonometric equations in the table

For sinus:
For cosine:
For Tangent and Kotnence:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

The solution of any trigonometric equation consists of two stages:

• by converting it to the simplest;
• to solve the resulting simplest equation, using the above written formulas of the roots and tables.

Consider the basic methods of solutions on the examples.

Algebraic method.

In this method, the variable is replaced and its substitution into equality.

Example. Solve equation: `2cos ^ 2 (X + \\ FRAC \\ PI 6) -3Sin (\\ FRAC \\ PI 3 - X) + 1 \u003d 0``

`2cos ^ 2 (X + \\ FRAC \\ PI 6) -3COS (X + \\ FRAC \\ PI 6) + 1 \u003d 0`

we make a replacement: `COS (X + \\ FRAC \\ PI 6) \u003d Y`, then` 2y ^ 2-3Y + 1 \u003d 0`,

we find the roots: `y_1 \u003d 1, y_2 \u003d 1/2, from which two cases follow:

1. `COS (X + \\ FRAC \\ PI 6) \u003d 1`,` X + \\ FRAC \\ pi 6 \u003d 2 \\ pi n`, `x_1 \u003d - \\ FRAC \\ PI 6 + 2 \\ PI N`.

2. `COS (X + \\ FRAC \\ PI 6) \u003d 1/2`,` X + \\ FRAC \\ PI 6 \u003d \\ PM Arccos 1/2 + 2 \\ pi n`, `x_2 \u003d \\ pm \\ frac \\ pi 3- \\ FRAC \\ PI 6 + 2 \\ PI N`.

Answer: `X_1 \u003d - \\ FRAC \\ PI 6 + 2 \\ PI n`,` x_2 \u003d \\ pm \\ frac \\ pi 3- \\ FRAC \\ pi 6 + 2 \\ pi n`.

Factorization.

Example. Solve equation: `sin x + cos x \u003d 1`.

Decision. Move left all members of the equality: `sin x + cos x-1 \u003d 0`. Using, we transform and decompose the left part:

`sin x - 2sin ^ 2 x / 2 \u003d 0`

`2sin x / 2 cos x / 2-2sin ^ 2 x / 2 \u003d 0`

`2sin x / 2 (cos x / 2-sin x / 2) \u003d 0`

1. `sin x / 2 \u003d 0`,` x / 2 \u003d \\ pi n`, `x_1 \u003d 2 \\ pi n`.
2. `cos x / 2-sin x / 2 \u003d 0,` tg x / 2 \u003d 1`, `x / 2 \u003d arctg 1+ \\ pi n`,` x / 2 \u003d \\ pi / 4 + \\ pi n` , `x_2 \u003d \\ pi / 2 + 2 \\ pi n`.

Answer: `x_1 \u003d 2 \\ pi n`,` x_2 \u003d \\ pi / 2 + 2 \\ pi n`.

Bringing to a homogeneous equation

Initially, this trigonometric equation should be brought to one of two types:

`a sin x + b cos x \u003d 0 (homogeneous equation of the first degree) or` a sin ^ 2 x + b sin x cos x + c cos ^ 2 x \u003d 0` (homogeneous equation of the second degree).

Then divide both parts on `COS X \\ NE 0` - for the first case, and on` cos ^ 2 x \\ ne 0` - for the second. We obtain the equation relative to TG X`: `a TG x + B \u003d 0 and` a TG ^ 2 x + b TG x + C \u003d 0`, which you need to solve well-known methods.

Example. Solve equation: `2 sin ^ 2 x + sin x cos x - cos ^ 2 x \u003d 1`.

Decision. We write the right side as `1 \u003d sin ^ 2 x + cos ^ 2 x`:

`2 sin ^ 2 x + sin x cos x - cos ^ 2 x \u003d` `sin ^ 2 x + cos ^ 2 x`,

`2 sin ^ 2 x + sin x cos x - cos ^ 2 x -`` sin ^ 2 x - cos ^ 2 x \u003d 0`

`sin ^ 2 x + sin x cos x - 2 cos ^ 2 x \u003d 0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right parts for `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) \u003d 0``

`Tg ^ 2 x + TG x - 2 \u003d 0`. We introduce the replacement `TG X \u003d T`, as a result of` T ^ 2 + T - 2 \u003d 0`. The roots of this equation: `T_1 \u003d -2` and` T_2 \u003d 1`. Then:

1. `TG x \u003d -2`,` x_1 \u003d arctg (-2) + \\ pi n`, `n \\ in z`
2. `TG x \u003d 1`,` x \u003d arctg 1+ \\ pi n`, `x_2 \u003d \\ pi / 4 + \\ pi n`,` n \\ in z`.

Answer. `x_1 \u003d arctg (-2) + \\ pi n`,` n \\ in z`, `x_2 \u003d \\ pi / 4 + \\ pi n`,` n \\ in z`.

Transition to a half-corner

Example. Solve Equation: `11 SIN X - 2 COS X \u003d 10`.

Decision. Applicate double angle formulas, as a result: `22 sin (x / 2) COS (X / 2) -`` 2 cos ^ 2 x / 2 + 2 sin ^ 2 x / 2 \u003d` `10 sin ^ 2 x / 2 +10 COS ^ 2 x / 2`

`4 TG ^ 2 x / 2 - 11 TG x / 2 + 6 \u003d 0`

Applying the algebraic method described above, we get:

1. `TG x / 2 \u003d 2`,` x_1 \u003d 2 arctg 2 + 2 \\ pi n`, `n \\ in z`,
2. `TG x / 2 \u003d 3/4`,` x_2 \u003d arctg 3/4 + 2 \\ pi n`, `n \\ in z`.

Answer. `x_1 \u003d 2 arctg 2 + 2 \\ pi n, n \\ in z`,` x_2 \u003d arctg 3/4 + 2 \\ pi n`, `n \\ in z`.

The introduction of auxiliary corner

In the trigonometric equation `a sin x + b cos x \u003d c`, where a, b, c - coefficients, and x is a variable, we divide both parts on` SQRT (A ^ 2 + B ^ 2) `:

`\\ FRAC A (SQRT (A ^ 2 + B ^ 2)) SIN X +` `\\ FRAC B (SQRT (A ^ 2 + B ^ 2)) COS X \u003d` `\\ FRAC C (SQRT (A ^ 2 + b ^ 2)) `.

The coefficients in the left part have the properties of the sinus and cosine, namely the sum of their squares equal to 1 and their modules are not more than 1. Denote them as follows: `\\ FRAC A (SQRT (A ^ 2 + B ^ 2)) \u003d COS \\ Varphi` , `\\ FRAC B (SQRT (A ^ 2 + B ^ 2)) \u003d SIN \\ Varphi`,` \\ FRAC C (SQRT (A ^ 2 + B ^ 2)) \u003d C`, then:

`COS \\ VARPHI SIN X + SIN \\ VARPHI COS X \u003d C`.

Let us consider in more detail on the following example:

Example. Solve equation: `3 sin x + 4 cos x \u003d 2`.

Decision. We divide both parts of equality on `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)) \u003d` `\\ FRAC 2 (SQRT (3 ^ 2 + 4 ^ 2)) `

`3/5 sin x + 4/5 cos x \u003d 2/5`.

Denote by `3/5 \u003d COS \\ Varphi`,` 4/5 \u003d sin \\ varphi`. Since `sin \\ varphi\u003e 0,` cos \\ varphi\u003e 0, then as an auxiliary angle, take `\\ varphi \u003d arcsin 4/5`. Then our equality will write in the form:

`COS \\ Varphi Sin X + SIN \\ Varphi COS X \u003d 2/5`

By applying the sum of the sum of the corners for sinus, we write out our equality in the following form:

`sin (x + \\ varphi) \u003d 2/5`

`X + \\ Varphi \u003d (- 1) ^ n Arcsin 2/5 + \\ PI N`,` n \\ in z`,

`X \u003d (- 1) ^ n Arcsin 2/5-` `arcsin 4/5 + \\ pi n`,` n \\ in z`.

Answer. `X \u003d (- 1) ^ n Arcsin 2/5-` `arcsin 4/5 + \\ pi n`,` n \\ in z`.

Fractional-rational trigonometric equations

These are equality with fractions, in the numerators and denominators of which there are trigonometric functions.

Example. Solve equation. `\\ FRAC (SIN X) (1 + COS X) \u003d 1-COS X`.

Decision. Multiply and divide the right-hand side of the equality on `(1 + cos x)`. As a result, we get:

`\\ FRAC (SIN X) (1 + COS X) \u003d` `\\ FRAC ((1-COS X) (1 + COS X)) (1 + COS x)`

`\\ FRAC (SIN X) (1 + COS X) \u003d` `\\ FRAC (1-COS ^ 2 x) (1 + COS x)`

`\\ FRAC (SIN X) (1 + COS X) \u003d` `\\ FRAC (SIN ^ 2 x) (1 + COS X)`

`\\ FRAC (SIN X) (1 + COS X) -`` \\ FRAC (SIN ^ 2 x) (1 + COS x) \u003d 0`

`\\ FRAC (SIN X-SIN ^ 2 x) (1 + COS x) \u003d 0`

Considering that the denominator is equal to being zero cannot, we get `1 + COS X \\ NE 0,` COS X \\ NE -1`, `x \\ ne \\ pi + 2 \\ pi n, n \\ in z`.

We equate to zero the numerator fraction: `sin x-sin ^ 2 x \u003d 0`,` sin x (1-sin x) \u003d 0`. Then `sin x \u003d 0` or` 1-sin x \u003d 0`.

1. `sin x \u003d 0`,` x \u003d \\ pi n`, `n \\ in z`
2. `1-sin x \u003d 0`,` sin x \u003d -1`, `x \u003d \\ pi / 2 + 2 \\ pi n, n \\ in z`.

Considering that `x \\ ne \\ pi + 2 \\ pi n, n \\ in z`, solutions will be` x \u003d 2 \\ pi n, n \\ in z` and `x \u003d \\ pi / 2 + 2 \\ pi n` , `n \\ in z`.

Answer. `x \u003d 2 \\ pi n`,` n \\ in z`, `x \u003d \\ pi / 2 + 2 \\ pi n`,` n \\ in z`.

Trigonometry, and trigonometric equations in particular, are used in almost all spheres of geometry, physics, engineering. Studying in the 10th grade begins, tasks are necessarily present for the exam, so try to remember all the formulas of trigonometric equations - they will definitely use you!

However, it is not necessary to remember them, the main thing is to understand the essence, and be able to withdraw. It is not difficult, as it seems. Be sure to watch the video.