Square equations with sine and cosine. Trigonometric equations

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The equation. Root of the equation. What does it mean to "solve the equation"?

The equation is equality containing a variable.

The root of the equation is the value of the variable, which, when substituting it into the equation, turns it into the correct numerical equality.

Solve the equation is to find all his roots or prove that there is no roots.

The system of equations is a combination of two or more equations with two and more unknown; Moreover, the solution of one of the equations is simultaneously by the solution of all others.

Types of equations and their solution: linear, square.

Linear equations - These are the equations of the form: ah + b \u003d 0, where a and b are some permanent. If not equal to zero, the equation has one single root: x \u003d - B: a. If a is zero and b is zero, then the root of the ah + b \u003d 0 equation is any number. If a is zero, and b is not zero, then the equation ah + b \u003d 0 does not have roots.

Methods for solving linear equations

1) identical transformations

2) graphic method.

Quadratic equation - this is the equation of type aX. 2 + bX. + c. \u003d 0, where coefficients a., b. and c. - arbitrary numbers, and a ≠ 0.

Let the square equation be given aX. 2 + bX. + c. \u003d 0. Then the discriminant is the number D. = b. 2 − 4aC.

1. If D. < 0, корней нет;

2. If D. \u003d 0, there is exactly one root;

3. If D. \u003e 0, the roots will be two.

If discriminant D\u003e 0, roots can be found by formulas: roots square equation. We now turn, actually, to the decision. If discriminant D. \u003e 0, roots can be found by formulas:

The solution of the simplest trigonometric equations

General view of the solution of the COS X \u003d A equation, where | a | ≤ 1, determined by the formula:

x \u003d ± arccos (a) + 2πk, k ∈ Z (integers), with | a | \u003e 1 COS X \u003d A equation has no solutions among real numbers.

General view of the solution SIN X \u003d A equation, where | a | ≤ 1, determined by the formula:

x \u003d (- 1) k · arcsin (a) + πk, k ∈ Z (integers), with | a | \u003e 1 The equation Sin X \u003d A has no solutions among real numbers.

The general type of solution of the equation TG X \u003d A is determined by the formula:

x \u003d arctg (a) + πk, k ∈ Z (integers).

The general view of the solution of the CTG X \u003d A equation is determined by the formula:

x \u003d arcctg (a) + πk, k ∈ Z (integers).

Solution of linear trigonometric equations

Linear trigonometric equations have the form K * f (x) + b \u003d 0, where F (x) is a trigonometric function, and k and b - valid numbers.

To solve the equation, it leads to the simplest type by identical transformations

Solution of linear - combined trigonometric equations

Linearly combined trigonometric equations have the form f (kx + b) \u003d a, where f (x) is a trigonometric function, a, k and b - valid numbers.

To solve the equation, it is introduced a new variable y \u003d kx + b. The resulting simplest trigonometric equation relative to y and produce reverse replacement.

Solving trigonometric equations using the formula

Solution of trigonometric equations using trigonometric identities

When solving trigonometric equations that are not simplest, identical transformations are carried out according to the following formulas:

Solution of square trigonometric equations

Distinctive features of equations reduced to square:

The equation contains trigonometric functions from one argument or they are easily reduced to one argument.

In the equation, there is only one trigonometric function or all functions can be reduced to one.

Algorithm Solutions:

The substitution is performed.

The expression transformation is performed.

An designation is entered (for example, SINX \u003d Y).

Square equation is solved.

The value of the designated value is substituted, and the trigonometric equation is solved

Moscow Department of Education

State budget professional

Educational institution of the city of Moscow

"Polytechnic Technical School No. 47 named after VG Fedorov"

Lesson

on discipline mathematics

"Trigonometric equations reduced to square"

Teacher

Protasevich Olga Nikolaevna

PROFESSION: Hardware and Software

DISCIPLINE : Mathematics

COURSE : 1

SEMESTER : 2

GROUP :

Theme lesson:

"Trigonometric equations reduced to square."

Type of lesson: combined lesson

Form of the lesson: Collective training according to the method of V.K. Dyachenko

(training in small groups)

Objectives lesson:

Education - consider general approaches, summarize information on the types and methods of solving trigonometric equations that are reduced to square; To form skills and skills to apply knowledge when solving basic equations and the use of knowledge gained in professional activities.

Developing - promote development logical thinking For students, develop the ability to analyze, reason, compare, draw conclusions, comprehend material;

Educational - Education of cognitive interest, elements of the culture of communication, encourage students for overcoming difficulties in the process of mental activity, the formation of work skills in the labor and training team.

Task lesson:

To acquaint trained with the main types and methods of solving trigonometric equations reduced to square.

Providing (resources):

Hardware: computer, multimedia projector.

Software:Microsoft.Excel.

Basic concepts:

Quadratic equation; the simplest trigonometric equations; inverse trigonometric functions; Trigonometric equations reduced to square.

Literature:

Bashmakov M.I. Mathematics: Tutorial for primary and medium vocational education.- m.; "Academy", 2010. - 256 p.

Dyachenko V. K. - M.; " Popular Education", 2001. - 496 p.

Methodical literature:

Bashmakov M.I. Mathematics: Book for teachers. Methodological manual. - M.; « Academy ", 2013- 224 p.

Electronic resources:

Site materials Social and pedagogical movement to create a collective method of training:www.kco-kras.ru.

Stages lesson

Organizing time.

Check your homework.

Actualization of reference knowledge.

Studying a new material.

Consolidation and systematization of knowledge gained.

Reflection. Summarizing. Homework.

During the classes

Organizing time.

The teacher puts the objectives of the lesson in front of the student:

1) become acquainted with the main types of trigonometric equations reduced to square;

2) introduction with typical methods of solving trigonometric equations reduced to square.

3) to teach apply the knowledge and skills to solve standard equations;

4) teach work with the information presented in various forms, carry out mutual control and self-control, apply the knowledge gained in professional activities.

II. . Check your homework.

The teacher includes a "homework" presentation, according to which the students independently check the homework, if necessary, make amendments and fixes to work.

At the request of the trained, the teacher comments on solutions to equations that caused difficulties, after which it announces the names of the students who, at the end of the lesson, gives to test the notebook.

№ 1

Answer:

№ 2

Answer:

№ 3

Answer:

№ 4

because then the root equation does not have

Answer: No roots

№ 5

Answer:

№ 6

Answer:

III . Actualization of reference knowledge.

The teacher forms educational groups / pairs and offers on issued forms to establish a correspondence between equations and answers: "You have a slide with a learning task. Install the correspondence between the equations (left part of the table) and the answers (right part of the table). Write down the numbers of faithful pairs of statements in the notebook. "

The specified tasks are duplicated on the presentation included.

Set compliance

p / P.

The equation

p / P.

Answer

No roots

At the end of the work, the teacher Frontally polls representatives of the groups, after which includes a presentation page with the correct solutions.

Right answers

p / P.

The equation

p / P.

Answer

No roots

11.

13.

10.

12.

IV . Studying a new material.

The teacher includes a presentation of a new material "Trigonometric equations reduced to square. Types of equations and methods of their solutions. "

Offers the learner to record the necessary theses and begins to comment on each slide, after which includes a presentation.

We introduce the concept:

General view of the square equation:

1 Type of trigonometric equations reduced to square - equations, algebraic relative to one of the trigonometric functions.

The teacher explains how to solve.

1. Direct substitution

Replacement ,

and

no roots

Answer:

A similar solution has a view equation

Replacement

2. Eurasions requiring conversion by the formula of a trigonometric unit

Replacement , then the equation takes the view

and

No roots

Answer:

A similar solution has the equation of the form:

replace , using a trigonometric unit formula

We obtain the equation containing only one trigonometric function :

Replacement

3. Eurasions requiring conversion by communication formula tGX. and from tGX.

We use the formula:

Multiply Equation on

Replacement , then the equation takes the view

and

Answer:

2 type trigonometric equations reduced to square- homogeneous equations in which each term has the same degree.

We divide the equation based on

Replacement , then the equation takes the view

and

Answer:

The teacher proposes to summarize the submitted material and asks questions: "How many types are trigonometric equations that are jammed to square? Their name? Name how to solve trigonometric equations that are reduced to square. "

The teacher sends the actions of the student in the preparation of the algorithm for solving the equations of this type.

Trigonometric equations reduced to square are divided into two main types:

tGX. and from tGX. :

2 Type - homogeneous equations in which each allegiate has the same degree:

The teacher is adjusted Algorithm Solutions:

1. Determine the type of equation. If necessary, convert the equation so that only one trigonometric function would be present in it. To do this, choose the desired formula: oror stripped by

2. Replacement is introduced (for example, sinx \u003d. t. , cOSX. = t. , tGX. = t. ).

5. Write the answer.

To secure the knowledge gained, the teacher proposes to establish a correspondence between the equations and possible methods of their solutions: "You have a slide with the study task.

1. Perform the classification of equations by decision methods according to the table below.

(Printed table options are on your tables).

2. Put a solution method number in the appropriate graph.

Fill the table".

Work is performed in pairs.

p / P.

The equation

method

Methods:

1) Enter a new variable.

2) Enter a new variable

3) Enter a new variable.

4) Convert the equation by applying the formula, enter a new variable.

5) Convert the equation by applying the formula, enter a new variable.

6) Divide each member of the equation on, enter a new variable.

7) Convert the equation by applying the formula, multiply the members of the equation on, enter a new variable.

Checking the task is carried out in the form of a frontal conversation.

Lecturer: "You have a slide with the right answers to the study task . Perform a check, referring to the correct answers to the learning task. Perform work on errors in the notebook. "

Blanks with tasks are collected at the end of the lesson.

p / P.

The equation

method

VI . Consolidation and systematization of knowledge gained.

The teacher offers learners to continue working in groups.

Lecturer: "Decide equations. Check the result in the editor Microsoft. Excel . At the end of the decision, the representative of the group goes to the educational board and represents the solution of the equation performed by the Group. " The teacher verifies the solution, assesses the work of the group and, if necessary, indicates errors. "

Teacher:

1 ) Discuss ways to solve in the group.

2) Write down the solution and the resulting answer to the notebook.

3) Perform the result check in the editor Microsoft. Excel .

4) Report readiness to the teacher.

5) Explain their decision by writing it on the board, members of other groups.

6) Thoughtfully listen to comrades' performances, ask questions if necessary.

Combat groups that fulfill the tasks in full, it is proposed to perform the task of other groups. The composition of successful groups is encouraged by an increase in the final score per unit.

First Group:

We use the formula:

and

No roots

because

Answer:

Second group:

We use the formula:

Replacement, then the equation takes the form

and

Answer:;

Third Group:

We use the formula:

Multiply Equation on

Replacement, then the equation takes the form

and

Answer:

Fourth group:

We divide the equation based on

Replacement, then the equation takes the form

and

Answer:

Fifth group:

Replacement, then the equation takes the form

and

Answer:; .

VII . Reflection. Summarizing. Homework.

Lecturer: Let's summarize your work, correlating the results of your activity with the goal.

Repeat concepts:

"Trigonometric equations that, with the help of conversion and replace the variable, are given to squares are called trigonometric equations that are reduced to square."

1 Type - equations, algebraic relative to one of the trigonometric functions:

- direct substitution - replacement or;

- equations requiring conversion by the formula of a trigonometric unit;

- equations requiring conversion by communication formula tGX. and S. tGX. :

2 Type - homogeneous equations in which each term has the same degree: we split the equation on, then replacing.

Algorithm Solutions:

1. Determine the type of equation. If necessary, convert the equation so that only one trigonometric function would be present in it.

To do this, choose the desired formula:

or or stripped by

2. Replacement is entered (for example, SINX \u003d t. , cOSX. = t. , tGX. = t. ).

3. Decide the square equation.

4. Reverse replacement is performed, and the simplest trigonometric equation is solved.

5. Write the answer.

The teacher is evaluating the work of trainees, training groups and announces evaluation.

Lecturer: "Write down homework: Bashmakov M.I. Mathematics: textbook for primary and secondary prof. Education. - M.; "Academy", 2010. Page 114-115. In the room 10, solve equations number 4,5,7,9. page 118. Perform a result check in the editor Microsoft. Excel ».

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. These tasks 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, to recall the necessary sequence of actions that will lead to the desired result. 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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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, there are no solutions among valid numbers.

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`

`sin x / 2 \u003d 0`,` x / 2 \u003d \\ pi n`, `x_1 \u003d 2 \\ pi n`.
`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:

`TG x \u003d -2`,` x_1 \u003d arctg (-2) + \\ pi n`, `n \\ in z`
`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:

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

`sin x \u003d 0`,` x \u003d \\ pi n`, `n \\ in z`
`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.

Lesson and presentation on the topic: "The solution of the simplest trigonometric equations"

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What we will study:
1. What is trigonometric equations?

3. Two basic methods for solving trigonometric equations.
4. Uniform trigonometric equations.
5. Examples.

What is trigonometric equations?

Guys, we have already studied Arksinus, Arkkosinus, Arctangent and Arkkothangence. Now let's look at trigonometric equations in general.

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

We repeat the type of solution of the simplest trigonometric equations:

1) If | A | ≤ 1, then the COS equation (x) \u003d a has a solution:

X \u003d ± Arccos (A) + 2πK

2) If | a | ≤ 1, then the equation sin (x) \u003d a has a solution:

3) if | A | \u003e 1, then the equation sin (x) \u003d a and cos (x) \u003d a do not have solutions 4) equation TG (x) \u003d A has a solution: x \u003d arctg (a) + πk

5) CTG equation (x) \u003d a has a solution: x \u003d ArcCTG (A) + πK

For all formulas k- integer

The simplest trigonometric equations are of the form: T (KX + M) \u003d A, T- any trigonometric function.

Example.

Solve equations: a) sin (3x) \u003d √3 / 2

Decision:

A) Denote 3x \u003d T, then our equation will rewrite in the form:

The solution to this equation will be: T \u003d ((- 1) ^ n) Arcsin (√3 / 2) + πn.

From the table of values, we obtain: T \u003d ((- 1) ^ n) × π / 3 + πn.

Let's go back to our variable: 3x \u003d ((- 1) ^ n) × π / 3 + πN,

Then X \u003d ((-1) ^ n) × π / 9 + πN / 3

Answer: X \u003d ((-1) ^ n) × π / 9 + πN / 3, where N-integer. (-1) ^ n - minus one to the degree n.

More examples of trigonometric equations.

Solve equations: a) cos (x / 5) \u003d 1 b) tg (3x- π / 3) \u003d √3

Decision:

A) This time we move directly to the calculation of the roots of the equation immediately:

X / 5 \u003d ± Arccos (1) + 2πk. Then x / 5 \u003d πk \u003d\u003e x \u003d 5πk

Answer: x \u003d 5πk, where k is an integer.

B) We write in the form: 3x- π / 3 \u003d arctg (√3) + πk. We know that: arctg (√3) \u003d π / 3

3x- π / 3 \u003d π / 3 + πk \u003d\u003e 3x \u003d 2π / 3 + πk \u003d\u003e x \u003d 2π / 9 + πk / 3

Answer: X \u003d 2π / 9 + πK / 3, where k is an integer.

Solve equations: COS (4x) \u003d √2 / 2. And find all the roots on the segment.

Decision:

Solving B. general Our equation: 4x \u003d ± Arccos (√2 \u200b\u200b/ 2) + 2πK

4x \u003d ± π / 4 + 2πk;

X \u003d ± π / 16 + πk / 2;

Now let's see what roots will fall on our segment. For k at k \u003d 0, x \u003d π / 16, we hit the specified segment.
At k \u003d 1, x \u003d π / 16 + π / 2 \u003d 9π / 16, they came again.
At k \u003d 2, x \u003d π / 16 + π \u003d 17π / 16, and here they did not get already, and therefore, with a large k, I also will not know.

Answer: x \u003d π / 16, x \u003d 9π / 16

Two main methods solutions.

We looked at the simplest trigonometric equations, but exist and more complex. To solve them, use the method of entering a new variable and the method of decomposition into multipliers. Let's consider examples.

Resolving equation:

Decision:
To solve our equation, we use the method of entering a new variable, we denote: T \u003d TG (X).

As a result of the replacement, we obtain: T 2 + 2T -1 \u003d 0

Find the roots of the square equation: T \u003d -1 and T \u003d 1/3

Then TG (x) \u003d - 1 and Tg (x) \u003d 1/3, they obtain the simplest trigonometric equation, we will find its roots.

X \u003d arctg (-1) + πk \u003d -π / 4 + πk; x \u003d arctg (1/3) + πk.

Answer: x \u003d -π / 4 + πk; x \u003d arctg (1/3) + πk.

Example of solving equation

Solve equations: 2Sin 2 (x) + 3 cos (x) \u003d 0

Decision:

We use the identity: SIN 2 (X) + COS 2 (X) \u003d 1

Our equation will take the form: 2-2COS 2 (x) + 3 cos (x) \u003d 0

2 cos 2 (x) - 3 cos (x) -2 \u003d 0

We introduce the replacement T \u003d COS (X): 2T 2 -3T - 2 \u003d 0

The solution of our square equation is the roots: T \u003d 2 and T \u003d -1 / 2

Then cos (x) \u003d 2 and cos (x) \u003d - 1/2.

Because Cosine can not accept values \u200b\u200bmore than one, then COS (X) \u003d 2 does not have roots.

For cos (x) \u003d - 1/2: x \u003d ± Arccos (-1/2) + 2πk; x \u003d ± 2π / 3 + 2πk

Answer: x \u003d ± 2π / 3 + 2πk

Uniform trigonometric equations.

Definition: The equation of the form A sin (x) + b cos (x) is called homogeneous trigonometric equations of the first degree.

View equations

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, we divide it on COS (X): You can not divide on the cosine if it is zero, let's make sure that it is not:
Let cos (x) \u003d 0, then asin (x) + 0 \u003d 0 \u003d\u003e sin (x) \u003d 0, but the sinus and cosine are not at the same time zero, they obtained a contradiction, so you can safely divide on zero.

Solve equation:
Example: COS 2 (X) + SIN (X) COS (x) \u003d 0

Decision:

I will summarize: COS (X) (C0S (X) + SIN (X)) \u003d 0

Then we need to solve two equations:

Cos (x) \u003d 0 and cos (x) + sin (x) \u003d 0

Cos (x) \u003d 0 at x \u003d π / 2 + πk;

Consider the COS equation (x) + sin (x) \u003d 0 we divide our equation on COS (X):

1 + TG (x) \u003d 0 \u003d\u003e TG (x) \u003d - 1 \u003d\u003e x \u003d arctg (-1) + πk \u003d -π / 4 + πk

Answer: x \u003d π / 2 + πk and x \u003d -π / 4 + πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, stick to these rules always!

1. To see what is equal to the coefficient A, if a \u003d 0 then, our equation will take the view COS (X) (BSIN (X) + CCOS (X)), an example of which the decision of which on the previous slide

2. If A ≠ 0, then you need to share both parts of the cosine equation in the square, we get:

We make the replacement of the variable T \u003d TG (X) we obtain the equation:

Solve example No: 3

Solve equation:
Decision:

We split both parts of the cosine equation Square:

We make the replacement of the variable T \u003d TG (X): T 2 + 2 T - 3 \u003d 0

Find the roots of the square equation: T \u003d -3 and T \u003d 1

Then: TG (x) \u003d - 3 \u003d\u003e x \u003d arctg (-3) + πk \u003d -arctg (3) + πk

TG (x) \u003d 1 \u003d\u003e x \u003d π / 4 + πk

Answer: X \u003d -ArCTG (3) + πk and x \u003d π / 4 + πk

Solve example No. 4

Solve equation:

Decision:
We transform our expression:

We can solve such equation: x \u003d - π / 4 + 2πk and x \u003d 5π / 4 + 2πk

Answer: x \u003d - π / 4 + 2πk and x \u003d 5π / 4 + 2πk

Solve example No: 5

Solve equation:

Decision:
We transform our expression:

We introduce the replacement TG (2x) \u003d T: 2 2 - 5T + 2 \u003d 0

The solution of our square equation will be the roots: T \u003d -2 and T \u003d 1/2

Then we get: TG (2x) \u003d - 2 and Tg (2x) \u003d 1/2
2x \u003d -ArcC (2) + πk \u003d\u003e x \u003d -arCTG (2) / 2 + πK / 2

2x \u003d arctg (1/2) + πk \u003d\u003e x \u003d arctg (1/2) / 2 + πk / 2

Answer: X \u003d -ArCTG (2) / 2 + πK / 2 and X \u003d ArCTG (1/2) / 2 + πK / 2

Tasks for self solutions.

1) solve equation

A) sin (7x) \u003d 1/2 b) cos (3x) \u003d √3 / 2 c) cos (-x) \u003d -1 g) tg (4x) \u003d √3 d) CTG (0.5x) \u003d -1.7

2) solve equations: sin (3x) \u003d √3 / 2. And find all the roots on the segment [π / 2; π].

3) solve equation: CTG 2 (x) + 2ctg (x) + 1 \u003d 0

4) solve equation: 3 sin 2 (x) + √3sin (x) cos (x) \u003d 0

5) Solve Equation: 3SIN 2 (3X) + 10 SIN (3X) COS (3X) + 3 COS 2 (3X) \u003d 0

6) Solve Equation: COS 2 (2X) -1 - COS (X) \u003d √3 / 2 -Sin 2 (2x)