Line UMK G. K. Muravin. Algebra and principles of mathematical analysis (10-11) (in-depth)

Line UMK G.K. Muravina, K.S. Muravina, O.V. Muravina. Algebra and principles of mathematical analysis (10-11) (basic)

How to teach solving trigonometric equations and inequalities: teaching methods

The mathematics course of the Russian Textbook Corporation, authored by Georgy Muravina and Olga Muravina, provides for a gradual transition to solving trigonometric equations and inequalities in the 10th grade, as well as continuing their study in the 11th grade. We present to your attention the stages of transition to the topic with excerpts from the textbook “Algebra and the beginning of mathematical analysis” (advanced level).

1. Sine and cosine of any angle (propaedeutic to the study of trigonometric equations)

Example assignment. Find approximately the angles whose cosines are equal to 0.8.

Solution. The cosine is the abscissa of the corresponding point on the unit circle. All points with abscissas equal to 0.8 belong to a straight line parallel to the ordinate axis and passing through the point C(0.8; 0). This line intersects the unit circle at two points: P α ° And P β ° , symmetrical about the abscissa axis.

Using a protractor we find that the angle α° approximately equal to 37°. So, the general view of the rotation angles with the end point P α°:

α° ≈ 37° + 360° n, Where n- any integer.

Due to symmetry about the abscissa axis, the point P β ° - end point of rotation at an angle of –37°. This means that for her the general form of rotation angles is:

β° ≈ –37° + 360° n, Where n- any integer.

Answer: 37° + 360° n, –37° + 360° n, Where n- any integer.

Example assignment. Find the angles whose sines are equal to 0.5.

Solution. The sine is the ordinate of the corresponding point on the unit circle. All points with ordinates equal to 0.5 belong to a straight line parallel to the abscissa axis and passing through the point D(0; 0,5).

This line intersects the unit circle at two points: Pφ and Pπ–φ, symmetrical about the ordinate axis. In a right triangle OKPφ leg KPφ is equal to half the hypotenuse OPφ , Means,

General view of rotation angles with end point P φ :

Where n- any integer. General view of rotation angles with end point P π–φ :

Where n- any integer.

Answer: Where n- any integer.

2. Tangent and cotangent of any angle (propaedeutics for the study of trigonometric equations)

Example 2.

Example assignment. Find the general form of angles whose tangent is –1.2.

Solution. Let us mark the point on the tangent axis C with an ordinate equal to –1.2, and draw a straight line O.C.. Straight O.C. intersects the unit circle at points P α ° And Pβ° - ends of the same diameter. The angles corresponding to these points differ from each other by an integer number of half-turns, i.e. 180° n (n- integer). Using a protractor we find that the angle P α° OP 0 is equal to –50°. This means that the general form of angles whose tangent is –1.2 is as follows: –50° + 180° n (n- integer)

Answer:–50° + 180° n, n∈ Z.

Using the sine and cosine of angles of 30°, 45° and 60°, it is easy to find their tangents and cotangents. For example,

The listed angles are quite common in various problems, so it is useful to remember the values ​​of the tangent and cotangent of these angles.

3. The simplest trigonometric equations

The following notations are introduced: arcsin α, arccos α, arctg α, arcctg α. It is not recommended to rush into introducing the combined formula. It is much more convenient to record two series of roots, especially when you need to select roots at intervals.

When studying the topic “the simplest trigonometric equations,” the equations are most often reduced to squares.

4. Reduction formulas

Reduction formulas are identities, i.e. they are true for any valid values φ . Analyzing the resulting table, you can see that:

1) the sign on the right side of the formula coincides with the sign of the reducible function in the corresponding quadrant, if we consider φ acute angle;

2) the name is changed only by the functions of the angles and

	φ + 2π n

5. Properties and graph of a function y = sin x

The simplest trigonometric inequalities can be solved either on a graph or on a circle. When solving a trigonometric inequality on a circle, it is important not to confuse which point to indicate first.

6. Properties and graph of a function y=cos x

The task of constructing a graph of a function y=cos x can be reduced to plotting the function y = sin x. Indeed, since graph of a function y=cos x can be obtained from the graph of the function y= sin x shifting the latter along the x-axis to the left by

7. Properties and graphs of functions y= tg x And y=ctg x

Function Domain y= tg x includes all numbers except numbers of the form where n ∈ Z. As when constructing a sinusoid, first we will try to obtain a graph of the function y = tg x in between

At the left end of this interval, the tangent is zero, and when approaching the right end, the tangent values ​​increase without limit. Graphically it looks like the graph of a function y = tg x presses against the straight line, going upward with it unlimitedly.

8. Dependencies between trigonometric functions of the same argument

Equality and express relations between trigonometric functions of the same argument φ. With their help, knowing the sine and cosine of a certain angle, you can find its tangent and cotangent. From these equalities it is easy to see that tangent and cotangent are related to each other by the following equality.

tg φ · cot φ = 1

There are other dependencies between trigonometric functions.

Equation of the unit circle centered at the origin x 2 + y 2= 1 connects the abscissa and ordinate of any point on this circle.

Fundamental trigonometric identity

cos 2 φ + sin 2 φ = 1

9. Sine and cosine of the sum and difference of two angles

Cosine sum formula

cos (α + β) = cos α cos β – sin α sin β

Difference cosine formula

cos (α – β) = cos α cos β + sin α sin β

Sine difference formula

sin (α – β) = sin α cos β – cos α sin β

Sine sum formula

sin (α + β) = sin α cos β + cos α sin β

10. Tangent of the sum and tangent of the difference of two angles

Tangent sum formula

Tangent difference formula

The textbook is included in the teaching materials in mathematics for grades 10–11 studying the subject at a basic level. Theoretical material is divided into mandatory and optional, the system of tasks is differentiated by level of difficulty, each chapter ends with test questions and assignments, and each chapter with a home test. The textbook includes project topics and links to Internet resources.

11. Trigonometric double angle functions

Double angle tangent formula

cos2α = 1 – 2sin 2 α cos2α = 2cos 2 α – 1

Example assignment. Solve the equation

Solution.

13. Solving trigonometric equations

In most cases, the original equation is reduced to simple trigonometric equations during the solution process. However, there is no single solution method for trigonometric equations. In each specific case, success depends on knowledge of trigonometric formulas and the ability to choose the right ones from them. However, the abundance of different formulas sometimes makes this choice quite difficult.

Equations that reduce to squares

Example assignment. Solve equation 2 cos 2 x+ 3 sin x = 0

Solution. Using the basic trigonometric identity, this equation can be reduced to a quadratic equation with respect to sin x:

2cos 2 x+3sin x= 0, 2(1 – sin 2 x) + 3sin x = 0,

2 – 2sin 2 x+3sin x= 0, 2sin 2 x– 3sin x – 2 = 0

Let's introduce a new variable y= sin x, then the equation will take the form: 2 y 2 – 3y – 2 = 0.

The roots of this equation y 1 = 2, y 2 = –0,5.

Returning to the variable x and we get the simplest trigonometric equations:

1) sin x= 2 – this equation has no roots, since sin x < 2 при любом значении x;

2) sin x = –0,5,

Answer:

Homogeneous trigonometric equations

Example assignment. Solve the equation 2sin 2 x– 3sin x cos x– 5cos 2 x = 0.

Solution. Let's consider two cases:

1)cos x= 0 and 2) cos x ≠ 0.

Case 1. If cos x= 0, then the equation takes the form 2sin 2 x= 0, whence sin x= 0. But this equality does not satisfy the cos condition x= 0, since under no circumstances x Cosine and sine do not vanish at the same time.

Case 2. If cos x≠ 0, then we can divide the equation by cos 2 x “Algebra and the beginning of mathematical analysis. 10th grade”, like many other publications, is available on the LECTA platform. To do this, take advantage of the offer.

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In this lesson we will look at basic trigonometric functions, their properties and graphs, and also list basic types of trigonometric equations and systems. In addition, we indicate general solutions of the simplest trigonometric equations and their special cases.

This lesson will help you prepare for one of the types of tasks B5 and C1.

Preparation for the Unified State Exam in mathematics

Experiment

Lesson 10. Trigonometric functions. Trigonometric equations and their systems.

Theory

Lesson summary

We have already used the term “trigonometric function” many times. Back in the first lesson of this topic, we defined them using a right triangle and a unit trigonometric circle. Using these methods of specifying trigonometric functions, we can already conclude that for them one value of the argument (or angle) corresponds to exactly one value of the function, i.e. we have the right to call sine, cosine, tangent and cotangent functions.

In this lesson, it's time to try to abstract from the previously discussed methods of calculating the values ​​of trigonometric functions. Today we will move on to the usual algebraic approach to working with functions, we will look at their properties and depict graphs.

Regarding the properties of trigonometric functions, special attention should be paid to:

The domain of definition and the range of values, because for sine and cosine there are restrictions on the range of values, and for tangent and cotangent there are restrictions on the range of definition;

The periodicity of all trigonometric functions, because We have already noted the presence of the smallest non-zero argument, the addition of which does not change the value of the function. This argument is called the period of the function and is denoted by the letter . For sine/cosine and tangent/cotangent these periods are different.

Consider the function:

1) Scope of definition;

2) Value range ;

3) The function is odd ;

Let's build a graph of the function. In this case, it is convenient to begin the construction with an image of the area that limits the graph from above with the number 1 and below with the number , which is associated with the range of values ​​of the function. In addition, for construction it is useful to remember the values ​​of the sines of several main table angles, for example, that this will allow you to build the first full “wave” of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will be repeated with an offset by a period, i.e. on .

Now let's look at the function:

The main properties of this function:

1) Scope of definition;

2) Value range ;

3) Even function This implies that the graph of the function is symmetrical about the ordinate;

4) The function is not monotonic throughout its entire domain of definition;

Let's build a graph of the function. As when constructing a sine, it is convenient to start with an image of the area that limits the graph at the top with the number 1 and at the bottom with the number , which is associated with the range of values ​​of the function. We will also plot the coordinates of several points on the graph, for which we need to remember the values ​​of the cosines of several main table angles, for example, that with the help of these points we can build the first full “wave” of the graph and then redraw it to the right and left, taking advantage of the fact that the picture will repeat with a period shift, i.e. on .

Let's move on to the function:

The main properties of this function:

1) Domain except , where . We have already indicated in previous lessons that it does not exist. This statement can be generalized by considering the tangent period;

2) Range of values, i.e. tangent values ​​are not limited;

3) The function is odd ;

4) The function increases monotonically within its so-called tangent branches, which we will now see in the figure;

5) The function is periodic with a period

Let's build a graph of the function. In this case, it is convenient to begin the construction by depicting the vertical asymptotes of the graph at points that are not included in the definition domain, i.e. etc. Next, we depict the tangent branches inside each of the strips formed by the asymptotes, pressing them to the left asymptote and to the right one. At the same time, do not forget that each branch increases monotonically. We depict all branches the same way, because the function has a period equal to . This can be seen from the fact that each branch is obtained by shifting the neighboring one along the abscissa axis.

And we finish with a look at the function:

The main properties of this function:

1) Domain except , where . From the table of values ​​of trigonometric functions, we already know that it does not exist. This statement can be generalized by considering the cotangent period;

2) Range of values, i.e. cotangent values ​​are not limited;

3) The function is odd ;

4) The function decreases monotonically within its branches, which are similar to the tangent branches;

5) The function is periodic with a period

Let's build a graph of the function. In this case, as for the tangent, it is convenient to begin the construction by depicting the vertical asymptotes of the graph at points that are not included in the definition area, i.e. etc. Next, we depict the branches of the cotangent inside each of the stripes formed by the asymptotes, pressing them to the left asymptote and to the right one. In this case, we take into account that each branch decreases monotonically. We depict all branches similarly to the tangent in the same way, because the function has a period equal to .

Separately, it should be noted that trigonometric functions with complex arguments may have a non-standard period. We are talking about functions of the form:

Their period is equal. And about the functions:

Their period is equal.

As you can see, to calculate a new period, the standard period is simply divided by the factor in the argument. It does not depend on other modifications of the function.

You can understand in more detail and understand where these formulas come from in the lesson about constructing and transforming graphs of functions.

We have come to one of the most important parts of the topic “Trigonometry”, which we will devote to solving trigonometric equations. The ability to solve such equations is important, for example, when describing oscillatory processes in physics. Let’s imagine that you have driven a few laps in a go-kart in a sports car; solving a trigonometric equation will help you determine how long you have been in the race depending on the position of the car on the track.

Let's write the simplest trigonometric equation:

The solution to such an equation is the arguments whose sine is equal to . But we already know that due to the periodicity of the sine, there is an infinite number of such arguments. Thus, the solution to this equation will be, etc. The same applies to solving any other simple trigonometric equation; there will be an infinite number of them.

Trigonometric equations are divided into several main types. Separately, we should dwell on the simplest ones, because everything else comes down to them. There are four such equations (according to the number of basic trigonometric functions). General solutions are known for them; they must be remembered.

The simplest trigonometric equations and their general solutions look like this:

Please note that the values ​​of sine and cosine must take into account the limitations known to us. If, for example, then the equation has no solutions and the specified formula should not be applied.

In addition, the specified root formulas contain a parameter in the form of an arbitrary integer. In the school curriculum, this is the only case when the solution to an equation without a parameter contains a parameter. This arbitrary integer shows that it is possible to write down an infinite number of roots of any of the above equations simply by substituting all the integers in turn.

You can get acquainted with the detailed derivation of these formulas by repeating the chapter “Trigonometric Equations” in the 10th grade algebra program.

Separately, it is necessary to pay attention to solving special cases of the simplest equations with sine and cosine. These equations look like:

Formulas for finding general solutions should not be applied to them. Such equations are most conveniently solved using the trigonometric circle, which gives a simpler result than general solution formulas.

For example, the solution to the equation is . Try to get this answer yourself and solve the remaining equations indicated.

In addition to the most common type of trigonometric equations indicated, there are several more standard ones. We list them taking into account those that we have already indicated:

1) Protozoa, For example, ;

2) Special cases of the simplest equations, For example, ;

3) Equations with complex argument, For example, ;

4) Equations reduced to their simplest by taking out a common factor, For example, ;

5) Equations reduced to their simplest by transforming trigonometric functions, For example, ;

6) Equations reduced to their simplest by substitution, For example, ;

7) Homogeneous equations, For example, ;

8) Equations that can be solved using the properties of functions, For example, . Don’t be alarmed by the fact that there are two variables in this equation; it solves itself;

As well as equations that are solved using various methods.

In addition to solving trigonometric equations, you must be able to solve their systems.

The most common types of systems are:

1) In which one of the equations is power, For example, ;

2) Systems of simple trigonometric equations, For example, .

In today's lesson we looked at the basic trigonometric functions, their properties and graphs. We also got acquainted with the general formulas for solving the simplest trigonometric equations, indicated the main types of such equations and their systems.

In the practical part of the lesson, we will examine methods for solving trigonometric equations and their systems.

Box 1.Solving special cases of the simplest trigonometric equations.

As we already said in the main part of the lesson, special cases of trigonometric equations with sine and cosine of the form:

have simpler solutions than those given by the general solution formulas.

A trigonometric circle is used for this. Let us analyze the method for solving them using the example of the equation.

Let us depict on the trigonometric circle the point at which the cosine value is zero, which is also the coordinate along the abscissa axis. As you can see, there are two such points. Our task is to indicate what the angle that corresponds to these points on the circle is equal to.

We start counting from the positive direction of the abscissa axis (cosine axis) and when setting the angle we get to the first depicted point, i.e. one solution would be this angle value. But we are still satisfied with the angle that corresponds to the second point. How to get into it?

Lesson and presentation on the topic: "Solving simple trigonometric equations"

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What we will 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 arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

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

Let us repeat the form of solving the simplest trigonometric equations:

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

X= ± arccos(a) + 2πk

2) If |a|≤ 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: T(kx+m)=a, T is some trigonometric function.

Example.

Solve the equations: a) sin(3x)= √3/2

Solution:

A) Let us 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 return 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 let’s move directly to calculating 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 it in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctan(√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 the equations: cos(4x)= √2/2. And find all the roots on the segment.

Solution:

Let us 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. At k At k=0, x= π/16, we are in the given segment.
With k=1, x= π/16+ π/2=9π/16, we hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means that for large k we also obviously won’t hit.

Answer: x= π/16, x= 9π/16

Two main solution methods.

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

Let's solve the equation:

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

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

Let's find the roots of the quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we get 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 will take the form: 2-2cos 2 (x) + 3 cos (x) = 0

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

Let us introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation is 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: Equations of the form a sin(x)+b cos(x) are 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, divide it by cos(x): You cannot divide by the cosine if it is equal to zero, let's make sure that this is not the case:
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 get a contradiction, so we can safely divide by zero.

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

Solution:

Let's 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 at 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, always follow these rules!

1. See what the coefficient a is equal to, if a=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 sides of the equation by the cosine squared, we get:

We change the variable t=tg(x) and get the equation:

Solve example No.:3

Solve the equation:
Solution:

Let's divide both sides of the equation by the cosine square:

We change the variable t=tg(x): t 2 + 2 t - 3 = 0

Let's find the roots of the quadratic equation: t=-3 and t=1

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

Tg(x)=1 => x= π/4+ πk

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

Solve example No.:4

Solve the equation:

Solution:
Let's transform our expression:

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

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

Solve example no.:5

Solve the equation:

Solution:
Let's transform our expression:

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

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

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

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

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

Problems for independent solution.

1) Solve the equation

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

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

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

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

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

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

You can order a detailed solution to your problem!!!

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 `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 simplest trigonometric equations are solved, as a rule, using formulas. Let me remind you that the simplest trigonometric equations are:

sinx = a

cosx = a

tgx = a

ctgx = a

x is the angle to be found,
a is any number.

And here are the formulas with which you can immediately write down the solutions to these simplest equations.

For sine:

For cosine:

x = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctan a + π n, n ∈ Z

For cotangent:

x = arcctg a + π n, n ∈ Z

Actually, this is the theoretical part of solving the simplest trigonometric equations. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off the charts. Especially if the example deviates slightly from the template. Why?

Yes, because a lot of people write down these letters, without understanding their meaning at all! He writes down with caution, lest something happen...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)

Let's figure it out?

One angle will be equal to arccos a, second: -arccos a.

And it will always work out this way. For any A.

If you don’t believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number A to something negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written as two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

x 2 = - arccos a + 2π n, n ∈ Z

Let's combine these two series into one:

x= ± arccos a + 2π n, n ∈ Z

And that's all. We have obtained a general formula for solving the simplest trigonometric equation with cosine.

If you understand that this is not some kind of superscientific wisdom, but just a shortened version of two series of answers, You will also be able to handle tasks “C”. With inequalities, with selecting roots from a given interval... There the answer with a plus/minus does not work. But if you treat the answer in a businesslike manner and break it down into two separate answers, everything will be resolved.) Actually, that’s why we’re looking into it. What, how and where.

In the simplest trigonometric equation

sinx = a

we also get two series of roots. Always. And these two series can also be recorded in one line. Only this line will be trickier:

x = (-1) n arcsin a + π n, n ∈ Z

But the essence remains the same. Mathematicians simply designed a formula to make one instead of two entries for series of roots. That's all!

Let's check the mathematicians? And you never know...)

In the previous lesson, the solution (without any formulas) of a trigonometric equation with sine was discussed in detail:

The answer resulted in two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

If we solve the same equation using the formula, we get the answer:

x = (-1) n arcsin 0.5 + π n, n ∈ Z

Actually, this is an unfinished answer.) The student must know that arcsin 0.5 = π /6. The complete answer would be:

x = (-1) n π /6+ π n, n ∈ Z

This raises an interesting question. Reply via x 1; x 2 (this is the correct answer!) and through lonely X (and this is the correct answer!) - are they the same thing or not? We'll find out now.)

We substitute in the answer with x 1 values n =0; 1; 2; etc., we count, we get a series of roots:

x 1 = π/6; 13π/6; 25π/6 and so on.

With the same substitution in response with x 2 , we get:

x 2 = 5π/6; 17π/6; 29π/6 and so on.

Now let's substitute the values n (0; 1; 2; 3; 4...) into the general formula for single X . That is, we raise minus one to the zero power, then to the first, second, etc. Well, of course, we substitute 0 into the second term; 1; 2 3; 4, etc. And we count. We get the series:

x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.

That's all you can see.) The general formula gives us exactly the same results as are the two answers separately. Just everything at once, in order. The mathematicians were not fooled.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we won’t.) They are already simple.

I wrote out all this substitution and checking specifically. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a short summary of the answers. For this brevity, we had to insert plus/minus into the cosine solution and (-1) n into the sine solution.

These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these insertions can easily unsettle a person.

So what should I do? Yes, either write the answer in two series, or solve the equation/inequality using the trigonometric circle. Then these insertions disappear and life becomes easier.)

We can summarize.

To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing down the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z

cosx = 0.2

No problem: x = ± arccos 0.2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctan 1,2 + π n, n ∈ Z

ctgx = 3.7

One left: x= arcctg3,7 + π n, n ∈ Z

cos x = 1.8

If you, shining with knowledge, instantly write the answer:

x= ± arccos 1.8 + 2π n, n ∈ Z

then you are already shining, this... that... from a puddle.) Correct answer: there are no solutions. Don't understand why? Read what arc cosine is. In addition, if on the right side of the original equation there are tabular values ​​of sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 and so on. - the answer through the arches will be unfinished. Arches must be converted to radians.

And if you come across inequality, like

then the answer is:

x πn, n ∈ Z

there is rare nonsense, yes...) Here you need to solve using the trigonometric circle. What we will do in the corresponding topic.

For those who heroically read to these lines. I simply cannot help but appreciate your titanic efforts. Bonus for you.)

Bonus:

When writing down formulas in an alarming combat situation, even seasoned nerds often get confused about where πn, And where 2π n. Here's a simple trick for you. In everyone formulas worth πn. Except for the only formula with arc cosine. It stands there 2πn. Two peen. Keyword - two. In this same formula there are two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign before the arc cosine, it’s easier to remember what will happen at the end two peen. And it also happens the other way around. The person will miss the sign ± , gets to the end, writes correctly two Pien, and he’ll come to his senses. There's something ahead two sign! The person will return to the beginning and correct the mistake! Like this.)

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By the way, I have a couple more interesting sites for you.)

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You can get acquainted with functions and derivatives.