Trigonometric functions of the equation. Basic methods for solving trigonometric equations

29.09.2019

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When solving many mathematical tasksEspecially those encountered up to 10 class, the procedure for actions performed, which will lead to the goal, is definitely defined. Such objectives include, for example, linear and square equations, linear and square inequalities, fractional equations and equations that are reduced to square. The principle of successful solution of each of the mentioned tasks is as follows: It is necessary to establish how the type is the solved task relates, to recall the necessary sequence of actions that will lead to the desired result, i.e. Answer, and perform these actions.

It is obvious that the success or failure in solving one or another task depends mainly on how correctly the type of equation is defined how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to own the skills of performing identical transformations and calculations.

Other situation is obtained with trigonometric equations. Establish the fact that the equation is trigonometric, absolutely not difficult. Difficulties appear when determining the sequence of actions that would led to the correct answer.

According to the appearance of the equation, sometimes it is difficult to determine its type. And not knowing the type of equation, it is almost impossible to choose from several dozen trigonometric formulas necessary.

To solve the trigonometric equation, you must try:

1. Create all functions included in the equation to the "same corners";
2. Create an equation to "identical functions";
3. Lay the left part of the factory equation, etc.

Consider basic methods for solving trigonometric equations.

I. Bringing to the simplest trigonometric equations

Schematic solution

Step 1. Express trigonometric function through well-known components.

Step 2. Find an argument function by formulas:

cos x \u003d a; x \u003d ± Arccos a + 2πn, n єz.

sin x \u003d a; x \u003d (-1) n arcsin a + πn, n є z.

tG X \u003d A; x \u003d arctg a + πn, n є z.

cTG X \u003d A; x \u003d arcctg a + πn, n є z.

Step 3. Find an unknown variable.

Example.

2 cos (3x - π / 4) \u003d -√2.

Decision.

1) COS (3X - π / 4) \u003d -√2 / 2.

2) 3x - π / 4 \u003d ± (π - π / 4) + 2πn, n є z;

3x - π / 4 \u003d ± 3π / 4 + 2πn, N є Z.

3) 3x \u003d ± 3π / 4 + π / 4 + 2πn, n є z;

x \u003d ± 3π / 12 + π / 12 + 2πn / 3, n є z;

x \u003d ± π / 4 + π / 12 + 2πn / 3, n є z.

Answer: ± π / 4 + π / 12 + 2πn / 3, n є z.

II. Replacing the variable

Schematic solution

Step 1. Create an equation to algebraic form relative to one of the trigonometric functions.

Step 2. Designate the resulting function of the variable T (if necessary, enter the restrictions on T).

Step 3. Record and solve the resulting algebraic equation.

Step 4. Make a replacement.

Step 5. Solve the simplest trigonometric equation.

Example.

2COS 2 (X / 2) - 5Sin (X / 2) - 5 \u003d 0.

Decision.

1) 2 (1 - sin 2 (x / 2)) - 5Sin (X / 2) - 5 \u003d 0;

2Sin 2 (X / 2) + 5Sin (X / 2) + 3 \u003d 0.

2) Let sin (x / 2) \u003d T, where | T | ≤ 1.

3) 2t 2 + 5t + 3 \u003d 0;

t \u003d 1 or E \u003d -3/2, does not satisfy the condition | T | ≤ 1.

4) sin (x / 2) \u003d 1.

5) x / 2 \u003d π / 2 + 2πn, n є z;

x \u003d π + 4πn, n є z.

Answer: x \u003d π + 4πn, n є z.

III. The method of lowering the order of the equation

Schematic solution

Step 1. Replace this linear equation using a degree reduction formula for this:

sin 2 x \u003d 1/2 · (1 - COS 2X);

cos 2 x \u003d 1/2 · (1 + cos 2x);

tG 2 X \u003d (1 - COS 2X) / (1 + COS 2x).

Step 2. Solve the obtained equation using methods I and II.

Example.

cOS 2X + COS 2 X \u003d 5/4.

Decision.

1) COS 2X + 1/2 · (1 + COS 2X) \u003d 5/4.

2) COS 2X + 1/2 + 1/2 · COS 2X \u003d 5/4;

3/2 · cos 2x \u003d 3/4;

2x \u003d ± π / 3 + 2πn, n є z;

x \u003d ± π / 6 + πn, n є z.

Answer: x \u003d ± π / 6 + πn, n є z.

IV. Uniform equations

Schematic solution

Step 1. Bring this equation to the form

a) a sin x + b cos x \u003d 0 (homogeneous equation of the first degree)

or to sight

b) a Sin 2 x + b sin x · cos x + c cos 2 x \u003d 0 (homogeneous equation of the second degree).

Step 2. Split both parts of the equation on

a) cos x ≠ 0;

b) COS 2 x ≠ 0;

and get the equation relative to TG X:

a) a tg x + b \u003d 0;

b) a TG 2 x + B arctg x + c \u003d 0.

Step 3. Solve equation by known methods.

Example.

5Sin 2 x + 3sin x · COS X - 4 \u003d 0.

Decision.

1) 5Sin 2 x + 3sin x · COS X - 4 (SIN 2 x + COS 2 x) \u003d 0;

5Sin 2 x + 3sin x · COS X - 4SIN² x - 4cos 2 x \u003d 0;

sIN 2 X + 3SIN X · COS X - 4COS 2 x \u003d 0 / COS 2 x ≠ 0.

2) TG 2 X + 3TG X - 4 \u003d 0.

3) Let TG x \u003d T, then

t 2 + 3T - 4 \u003d 0;

t \u003d 1 or t \u003d -4, then

tG x \u003d 1 or TG x \u003d -4.

From the first equation x \u003d π / 4 + πn, n є z; From the second equation x \u003d -arctg 4 + πk, k є z.

Answer: x \u003d π / 4 + πn, n є z; x \u003d -arctg 4 + πk, k є z.

V. Method of converting an equation using trigonometric formulas

Schematic solution

Step 1. Using all sorts of trigonometric formulas, lead this equation to the equation, solved methods I, II, III, IV.

Step 2. Solve the resulting equation known methods.

Example.

sIN X + SIN 2X + SIN 3X \u003d 0.

Decision.

1) (SIN X + SIN 3X) + SIN 2X \u003d 0;

2Sin 2X · COS X + SIN 2X \u003d 0.

2) sIN 2X · (2cos x + 1) \u003d 0;

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

From the first equation 2x \u003d π / 2 + πn, n є z; From the second equation COS X \u003d -1/2.

We have x \u003d π / 4 + πn / 2, n є z; From the second equation x \u003d ± (π - π / 3) + 2πk, k є z.

As a result, x \u003d π / 4 + πn / 2, n є z; x \u003d ± 2π / 3 + 2πk, k є z.

Answer: x \u003d π / 4 + πn / 2, n є z; x \u003d ± 2π / 3 + 2πk, k є z.

Skills and skills to solve trigonometric equations are very important, their development requires considerable efforts, both by the student and from the teacher.

With the solution of trigonometric equations, many challenges of stereometry, physics, and others are associated with the process of solving such tasks, as it were, concludes many knowledge and skills, which are purchased in the study of elements of trigonometry.

Trigonometric equations occupy an important place in the process of learning mathematics and personality development as a whole.

Have questions? Do not know how to solve trigonometric equations?
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It requires knowledge of the basic formulas of trigonometry - the sum of the squares of sinus and cosine, the expression of the tangent through sinus and cosine and others. For those who forgot them or does not know, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations With the right approach, a fairly exciting activity, such as, for example, to collect Rubik's cube.

Based on the very name, it can be seen that the trigonometric equation is an equation in which the unknown is under a trigonometric function.
There are so-called simple trigonometric equations. Here's what they look: sinh \u003d a, cos x \u003d a, tg x \u003d a. Consider how to solve such trigonometric equationsFor clarity, we will use the already familiar trigonometric circle.

sinh \u003d a.

cOS X \u003d A

tG X \u003d A

cot X \u003d A

Any trigonometric equation is solved in two stages: give the equation to the simplest form and then solve it as the simplest trigonometric equation.
There are 7 basic methods with which trigonometric equations are solved.

Method for replacing a variable and substitution

Solve equation 2COS 2 (x + / 6) - 3sin (/ 3 - x) +1 \u003d 0

Using the formulas, we get:

2COS 2 (X + / 6) - 3COS (X + / 6) +1 \u003d 0

Replace COS (X + / 6) to Y to simplify and get a conventional square equation:

2y 2 - 3Y + 1 + 0

The roots of which y 1 \u003d 1, y 2 \u003d 1/2

Now we go in reverse order

We substitute the found values \u200b\u200bof y and get two answers:

Solving trigonometric equations through multipliers decomposition

How to solve the equation sin x + cos x \u003d 1?

We transfer everything to the left to the right remains 0:

sIN X + COS X - 1 \u003d 0

We use the elevated identities to simplify the equation:

sIN X - 2 SIN 2 (X / 2) \u003d 0

We make expansion of multipliers:

2Sin (X / 2) * COS (X / 2) - 2 SIN 2 (X / 2) \u003d 0

2Sin (x / 2) * \u003d 0

We get two equations

Bringing to a homogeneous equation

The equation is homogeneous relative to sinus and cosine, if all its members relative to the sinus and cosine of the same degree of the same angle. To solve a homogeneous equation, enter as follows:

a) transfer all its members to the left side;

b) make all common factors for brackets;

c) equal all multipliers and brackets to 0;

d) in brackets obtained a homogeneous equation to a lesser extent, it in turn is divided into sinus or cosine to the high degree;

e) solve the resulting equation relative to TG.

Solve equation 3Sin 2 x + 4 SIN X COS X + 5 COS 2 x \u003d 2

We use the sin 2 x + cos 2 x \u003d 1 formula and get rid of the open twice to the right:

3Sin 2 x + 4 SIN X COS X + 5 COS X \u003d 2SIN 2 X + 2COS 2 x

sIN 2 X + 4 SIN X COS X + 3 COS 2 x \u003d 0

We divide on COS X:

tG 2 x + 4 TG x + 3 \u003d 0

We replace TG X to Y and we get a square equation:

y 2 + 4Y +3 \u003d 0, the roots of which y 1 \u003d 1, y 2 \u003d 3

From here we find two solutions of the source equation:

x 2 \u003d arctg 3 + k

Solving equations, through the transition to half corner

Solve equation 3SIN X - 5COS X \u003d 7

Go to X / 2:

6Sin (X / 2) * COS (X / 2) - 5COS 2 (X / 2) + 5Sin 2 (x / 2) \u003d 7Sin 2 (X / 2) + 7COS 2 (X / 2)

Preen all left:

2Sin 2 (X / 2) - 6Sin \u200b\u200b(X / 2) * COS (X / 2) + 12COS 2 (X / 2) \u003d 0

We divide on COS (X / 2):

tG 2 (X / 2) - 3TG (X / 2) + 6 \u003d 0

The introduction of auxiliary corner

For consideration, take the equation of the form: A SIN X + B COS X \u003d C,

where a, b, c is some arbitrary coefficients, and X is unknown.

Both parts of the equation are divided into:

Now the coefficients of the equation according to trigonometric formulas have the properties of SIN and COS, namely: their module is not more than 1 and the sum of the squares \u003d 1. Denote them, respectively, as COS and SIN, where it is the so-called auxiliary angle. Then the equation will take the form:

cOS * SIN X + SIN * COS X \u003d C

or sin (x +) \u003d c

By the solution of this simplest trigonometric equation will be

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

It should be noted that the designations of COS and SIN are interchangeable.

Solve sin 3x equation - COS 3X \u003d 1

In this equation, the coefficients:

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

The solution of the simplest trigonometric equations.

The solution of trigonometric equations of any level of complexity is ultimately reduced to solving the simplest trigonometric equations. And in this the best assistant again turns out to be a trigonometric circle.

Recall the definition of cosine and sinus.

The cosine of the angle is the abscissa (that is, the coordinate on the axis) of the point on the unit circle corresponding to the rotation at the given angle.

The sinus of the angle is called the ordinate (that is, the coordinate along the axis) of the point on the unit circle corresponding to the rotation at the given angle.

A positive direction of movement on a trigonometric circle is the movement counterclockwise. Turn on 0 degrees (or 0 radian) corresponds to a point with coordinates (1; 0)

We use these definitions to solve the simplest trigonometric equations.

1. Resolving equation

This equation satisfy all such values \u200b\u200bof the angle of rotation that correspond to the points of the circle, the ordinate of which is equal to.

We note on the ordinate axis the point with the ordinate:

We carry out the horizontal line parallel to the abscissa axis to the intersection with the circle. We will receive two points lying on the circle and having the ordinate. These points correspond to the angles of rotation and radians:

If we, coming out of the point corresponding to the angle of rotation on the radian, wage the full circle, then we will come to a point corresponding to the angle of rotation on the radian and having the same ordinate. That is, this corner of rotation also satisfies our equation. We can do how much "idle" revolutions, returning to the same point, and all these angles will satisfy our equation. The number of "idle" revolutions will indicate the letter (or). Since we can make these revs both in positive and in the negative direction, (or) can take any integer values.

That is, the first series of solutions of the source equation has the form:

, - Many integers (1)

Similarly, the second series of solutions has the form:

where,. (2)

As you guessed, the point of the circle is based on this series of solutions, corresponding to the angle of rotation on.

These two series of solutions can be combined into one entry:

If we take in this record (that is, even), then we will get the first series of solutions.

If we take in this record (that is, an odd), then we will get the second series of solutions.

2. Now let's solve equation

Since it is the abscissa of the point of a single circle obtained by turning to the angle, we note on the axis point with the abscissa:

We carry out the vertical line parallel to the axis to the intersection with the circle. We will get two points lying on the circle and having abscissa. These points correspond to the angles of rotation and radians. Recall that when moving clockwise, we get a negative angle of rotation:

We write two series of solutions:

(We fall at the desired point, passing out of the main full circle, that is.

We combine these two series in one entry:

3. Resolving the equation

The line of tangents passes through a point with coordinates (1.0) of a single circle parallel to the Oy axis

We note the point on it, with the ordinate of equal to 1 (we are looking for, the tangent of which angles is 1):

Connect this point with the start of the coordinates of the straight line and we note the intersection points of the straight line with a single circle. The intersection points of the direct and circle correspond to the corners of the turn on and:

Since the points corresponding to the angles of rotation that satisfy our equation lie at the distance of the radian from each other, then we can write the solution in this way:

4. Resolving equation

The catangens line passes through the point with the coordinates of a single circle parallel to the axis.

Note on the line of the catangents, the point with the abscissa -1:

Connect this point with the start of the coordinates direct and continue it to cross the circle. This direct cross the circle at points corresponding to the angles of rotation and radians:

Since these points will take apart from each other, equal, then the general solution of this equation we can write this way:

In the above examples illustrating the solution of the simplest trigonometric equations, table values \u200b\u200bof trigonometric functions were used.

However, if it is not a table value in the right part of the equation, then we substitute the value in the general solution of the equation: