Homogeneous trigonometric equations: general solution scheme. Solution of homogeneous trigonometric equations

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Nonlinear equations in two unknowns

linear

Systems of two equations, one of which is homogeneous

Examples of solving systems of equations of other types

Homogeneous trigonometric equation of the first degree

First example. Solution of a homogeneous trigonometric equation of the first degree

Homogeneous trigonometric equation of the second degree

Second example. Solution of a homogeneous trigonometric equation of the second degree

Third example. Solution of a homogeneous trigonometric equation of the second degree

Training tasks

Money

Lesson type: explanation of new material. Work takes place in groups. Each group has an expert who supervises and directs the work of the students. Helps weak students to believe in their strength in solving these equations.

Related lesson

" Homogeneous trigonometric equations"

(10th grade)

Target:

introduce the concept of homogeneous trigonometric equations of I and II degrees;
formulate and work out an algorithm for solving homogeneous trigonometric equations of I and II degrees;
teach students to solve homogeneous trigonometric equations of I and II degrees;
develop the ability to identify patterns, generalize;
stimulate interest in the subject, develop a sense of solidarity and healthy rivalry.

Lesson type : a lesson in the formation of new knowledge.

Conduct form: work in groups.

Equipment: computer, multimedia installation

During the classes

I. Organizational moment

In the lesson, a rating system for assessing knowledge (the teacher explains the system for assessing knowledge, filling out the assessment sheet by an independent expert selected by the teacher from among the students). The lesson is accompanied by a presentation. Attachment 1.

Evaluation sheet No.

II. Updating of basic knowledge..

We continue our study of the topic “Trigonometric Equations”. Today in the lesson we will get to know you with another type of trigonometric equations and methods for solving them, and therefore we will repeat what we have learned. All types of trigonometric equations when solved are reduced to solving the simplest trigonometric equations. Let us recall the main types of the simplest trigonometric equations. Use the arrows to match the expressions.

III. Motivation for learning.

We have to work on solving a crossword puzzle. Having solved it, we will learn the name of a new type of equations that we will learn to solve today in the lesson.

Questions are projected onto the board. Students guess, an independent expert enters points on the score sheet for the students who answer.

Having solved the crossword puzzle, the guys will read the word “homogeneous”.

Crossword.

If you enter the correct words, you get the name of one of the types of trigonometric equations.

1. The value of the variable that turns the equation into a true equality? (Root)

2. Unit of measure for angles? (Radian)

3. Numerical multiplier in the product? (Coefficient)

4. A section of mathematics that studies trigonometric functions? (Trigonometry)

5. What mathematical model is needed to introduce trigonometric functions? (Circle)

6. Which of the trigonometric functions is even? (Cosine)

7. What is the name of the true equality? (Identity)

8.Equality with a variable? (The equation)

9. Equations with the same roots? (equivalent)

10. Set of roots of the equation? (Solution)

IV. Explanation of new material.

The topic of the lesson is “Homogeneous trigonometric equations”. (Presentation)

Examples:

sin x + cos x = 0
√3cos x + sin x = 0
sin4x = cos4x
2sin 2 x + 3 sin x cos x + cos 2 x = 0
4 sin 2 x – 5 sin x cos x – 6 cos 2 x = 0
sin 2 x + 2 sin x cos x - 3 cos 2 x + 2 = 0
4sin 2 x – 8 sin x cos x + 10 cos 2 x = 3
1 + 7cos 2 x = 3 sin 2x
sin2x + 2cos2x = 1

V. Independent work

Tasks: to comprehensively test the knowledge of students when solving all types of trigonometric equations, to encourage students to introspection, self-control.
Students are asked to complete 10 minutes of written work.
Students perform on blank sheets of paper for copying. After the time has passed, tops of independent work are collected, and the solutions for copying remain with the students.
Checking independent work (3 min) is carried out by mutual checking.
. Students check the written work of their neighbor with a colored pen and write down the name of the verifier. Then hand over the leaves.

Then they are handed over to an independent expert.

Option 1: 1) sin x = √3cos x

2) 3sin 2 x - 7sin x cos x + 2 cos 2 x = 0

3) 3sin x – 2sin x cos x = 1

4) sin2x⁄sinx=0

Option 2: 1) cosx + √3sin x = 0

2)2sin 2 x + 3sin x cos x – 2 cos 2 x = 0

3)1 + sin 2 x = 2 sin x cos x

4) cos 2x ⁄ cos x = 0

VI. Summing up the lesson

VII. Homework:

Homework - 12 points (3 equations 4 x 3 = 12 were given for homework)

Student activity - 1 answer - 1 point (4 points maximum)

Solving Equations 1 point

Independent work - 4 points

Definition 1 . Let A be some set of pairs of numbers (x; y) . It is said that the set A is given numeric function z from two variables x and y , if a rule is specified, by which each pair of numbers from the set A is assigned a certain number.

Specifying a numerical function z of two variables x and y is often designate So:

where f (x , y) - any function other than the function

f (x , y) = ax+by+c ,

where a, b, c are given numbers.

Definition 3 . Equation (2) solution name a pair of numbers x; y) , for which formula (2) is a true equality.

Example 1 . solve the equation

Since the square of any number is non-negative, it follows from formula (4) that the unknowns x and y satisfy the system of equations

the solution of which is a pair of numbers (6 ; 3) .

Answer: (6; 3)

Example 2 . solve the equation

Therefore, the solution to equation (6) is an infinite number of pairs of numbers kind

(1 + y ; y) ,

where y is any number.

Definition 4 . Solving the system of equations

name a pair of numbers x; y) , substituting them into each of the equations of this system, we obtain the correct equality.

Systems of two equations, one of which is linear, have the form

g(x , y)

Example 4 . Solve a system of equations

Solution . Let us express the unknown y in terms of the unknown x from the first equation of system (7) and substitute the resulting expression into the second equation of the system:

Solving the Equation

x 1 = - 1 , x 2 = 9 .

Consequently,

y 1 = 8 - x 1 = 9 ,
y 2 = 8 - x 2 = - 1 .

Systems of two equations, one of which is homogeneous, have the form

where a , b , c are given numbers, and g(x , y) is a function of two variables x and y .

Example 6 . Solve a system of equations

Solution . Let's solve the homogeneous equation

3x 2 + 2xy - y 2 = 0 ,

3x 2 + 17xy + 10y 2 = 0 ,

treating it as a quadratic equation with respect to the unknown x:

In case when x = - 5y, from the second equation of system (11) we obtain the equation

5y 2 = - 20 ,

which has no roots.

In case when

from the second equation of system (11) we obtain the equation

whose roots are numbers y 1 = 3 , y 2 = - 3 . Finding for each of these values y the corresponding value x , we obtain two solutions to the system: (- 2 ; 3) , (2 ; - 3) .

Answer: (- 2 ; 3) , (2 ; - 3)

Example 8 . Solve the system of equations (MIPT)

Solution . We introduce new unknowns u and v , which are expressed in terms of x and y by the formulas:

To rewrite system (12) in terms of new unknowns, we first express the unknowns x and y in terms of u and v . It follows from system (13) that

We solve the linear system (14) by excluding the variable x from the second equation of this system. To this end, we perform the following transformations on system (14):

we leave the first equation of the system unchanged;
subtract the first equation from the second equation and replace the second equation of the system with the resulting difference.

As a result, system (14) is transformed into an equivalent system

from which we find

Using formulas (13) and (15), we rewrite the original system (12) as

The first equation of system (16) is linear, so we can express the unknown u from it in terms of the unknown v and substitute this expression into the second equation of the system.

With the help of this video lesson, students will be able to study the topic of homogeneous trigonometric equations.

Let's give definitions:

1) a homogeneous trigonometric equation of the first degree looks like a sin x + b cos x = 0;

2) a homogeneous trigonometric equation of the second degree looks like a sin 2 x + b sin x cos x + c cos 2 x = 0.

Consider the equation a sin x + b cos x = 0. If a is zero, then the equation will look like b cos x = 0; if b is zero, then the equation will look like a sin x = 0. These are the equations that we called the simplest and solved earlier in previous topics.

Now consider the option when a and b are not equal to zero. By dividing the parts of the equation by the cosine x and we will carry out the transformation. We get a tg x + b = 0, then tg x will be equal to - b/a.

From the above it follows that the equation a sin mx + b cos mx = 0 is a homogeneous trigonometric equation of the first degree. To solve an equation, divide its parts by cos mx.

Let's analyze example 1. Solve 7 sin (x / 2) - 5 cos (x / 2) = 0. First, divide the parts of the equation by cosine (x / 2). Knowing that the sine divided by the cosine is the tangent, we get 7 tg (x / 2) - 5 = 0. Transforming the expression, we find that the value of the tangent (x / 2) is 5/7. The solution to this equation is x = arctan a + πn, in our case x = 2 arctan (5/7) + 2πn.

Consider the equation a sin 2 x + b sin x cos x + c cos 2 x = 0:

1) with a equal to zero, the equation will look like b sin x cos x + c cos 2 x = 0. Transforming, we get the expression cos x (b sin x + c cos x) = 0 and proceed to the solution of two equations. After dividing the parts of the equation by the cosine x, we get b tg x + c = 0, which means tg x = - c/b. Knowing that x \u003d arctan a + πn, then the solution in this case will be x \u003d arctg (- c / b) + πn.

2) if a is not equal to zero, then, by dividing the parts of the equation by the cosine squared, we obtain an equation containing a tangent, which will be square. This equation can be solved by introducing a new variable.

3) when c equals zero, the equation will take the form a sin 2 x + b sin x cos x = 0. This equation can be solved by taking the sine of x out of the bracket.

1. see if there is a sin 2 x in the equation;

2. if the term a sin 2 x is contained in the equation, then the equation can be solved by dividing both parts by the cosine squared and then introducing a new variable.

3. if the equation a sin 2 x does not contain, then the equation can be solved by taking out the brackets cosx.

Consider example 2. We take out the cosine and get two equations. The root of the first equation is x = π/2 + πn. To solve the second equation, we divide the parts of this equation by the cosine x, by means of transformations we obtain x = π/3 + πn. Answer: x = π/2 + πn and x = π/3 + πn.

Let's solve example 3, an equation of the form 3 sin 2 2x - 2 sin 2x cos 2x + 3 cos 2 2x = 2 and find its roots that belong to the segment from - π to π. Because Since this equation is non-homogeneous, it is necessary to reduce it to a homogeneous form. Using the formula sin 2 x + cos 2 x = 1, we get the equation sin 2 2x - 2 sin 2x cos 2x + cos 2 2x = 0. Dividing all parts of the equation by cos 2 x, we get tg 2 2x + 2tg 2x + 1 = 0 Using the introduction of a new variable z = tg 2x, we solve the equation whose root is z = 1. Then tg 2x = 1, which implies that x = π/8 + (πn)/2. Because according to the condition of the problem, you need to find the roots that belong to the segment from - π to π, the solution will look like - π< x <π. Подставляя найденное значение x в данное выражение и преобразовывая его, получим - 2,25 < n < 1,75. Т.к. n - это целые числа, то решению уравнения удовлетворяют значения n: - 2; - 1; 0; 1. При этих значениях n получим корни решения исходного уравнения: x = (- 7π)/8, x = (- 3π)/8, x =π/8, x = 5π/8.

TEXT INTERPRETATION:

Homogeneous trigonometric equations

Today we will analyze how the "Homogeneous Trigonometric Equations" are solved. These are equations of a special kind.

Let's get acquainted with the definition.

Type equation and sinx+bcosx = 0 (and sine x plus be cosine x is zero) is called a homogeneous trigonometric equation of the first degree;

equation of the form a sin 2 x+bsin xcosx+ccos 2 x= 0 (and sine square x plus be sine x cosine x plus se cosine square x equals zero) is called a homogeneous trigonometric equation of the second degree.

If a=0, then the equation will take the form bcosx = 0.

If b = 0 , then we get and sin x = 0.

These equations are elementary trigonometric, and we considered their solution in our previous topics

Consider the case when both coefficients are non-zero. Divide both sides of the equation butsinx+ bcosx = 0 term by term on cosx.

We can do this, since cosine x is non-zero. After all, if cosx = 0 , then the equation butsinx+ bcosx = 0 will take the form butsinx = 0 , but≠ 0, therefore sinx = 0 . Which is impossible, because according to the basic trigonometric identity sin 2x+cos 2 x=1 .

Dividing both sides of the equation butsinx+ bcosx = 0 term by term on cosx, we get: + =0

Let's make the transformations:

1. Since = tg x, then =and tg x

2 reduce by cosx, then

Thus we get the following expression and tg x + b =0.

Let's do the transformation:

1. Move b to the right side of the expression with the opposite sign

and tg x \u003d - b

2. Get rid of the multiplier and dividing both sides of the equation by a

tg x= -.

Conclusion: An equation of the form and sinmx+bcosmx = 0 (and the sine em x plus the cosine em x is zero) is also called a homogeneous trigonometric equation of the first degree. To solve it, divide both sides by cosmx.

EXAMPLE 1. Solve the equation 7 sin - 5 cos \u003d 0 (seven sine x by two minus five cosine x by two is zero)

Solution. We divide both parts of the equation term by term by cos, we get

1. \u003d 7 tg (since the ratio of sine to cosine is tangent, then seven sine x by two divided by cosine x by two is equal to 7 tangent x by two)

2. -5 = -5 (when abbreviated cos)

Thus we got the equation

7tg - 5 = 0, Let's transform the expression, move minus five to the right side, changing the sign.

We have reduced the equation to the form tg t = a, where t=, a =. And since this equation has a solution for any value but and these solutions look like

x \u003d arctg a + πn, then the solution to our equation will look like:

Arctg + πn, find x

x \u003d 2 arctan + 2πn.

Answer: x \u003d 2 arctg + 2πn.

Let's move on to a homogeneous trigonometric equation of the second degree

butsin 2 x+b sin x cos x +fromcos2 x= 0.

Let's consider several cases.

I. If a=0, then the equation will take the form bsinxcosx+ccos 2 x= 0.

When solving e then we use the factorization method of the equations. Let's take out cosx brackets and we get: cosx(bsinx+ccosx)= 0 . Where cosx= 0 or

b sin x +fromcos x= 0. And we already know how to solve these equations.

We divide both parts of the equation term by term by cosx, we get

1 (because the ratio of sine to cosine is the tangent).

Thus we get the equation: b tg x+c=0

We have reduced the equation to the form tg t = a, where t= x, a =. And since this equation has a solution for any value but and these solutions look like

x \u003d arctg a + πn, then the solution to our equation will be:

x \u003d arctg + πn, .

II. If a≠0, then we divide both parts of the equation term by term into cos 2 x.

(Arguing similarly, as in the case of a homogeneous trigonometric equation of the first degree, cosine x cannot vanish).

III. If c=0, then the equation will take the form butsin 2 x+ bsinxcosx= 0. This equation is solved by the factorization method (take out sinx for brackets).

So, when solving the equation butsin 2 x+ bsinxcosx+ccos 2 x= 0 you can follow the algorithm:

EXAMPLE 2. Solve the equation sinxcosx - cos 2 x= 0 (sine x times cosine x minus the root of three times cosine squared x equals zero).

Solution. Let us factorize (bracket cosx). Get

cos x(sin x - cos x)= 0, i.e. cos x=0 or sin x - cos x= 0.

Answer: x \u003d + πn, x \u003d + πn.

EXAMPLE 3. Solve the equation 3sin 2 2x - 2 sin2xcos2 x +3cos 2 2x= 2 (three sine square of two x minus twice the product of the sine of two x and the cosine of two x plus three cosine square of two x) and find its roots belonging to the interval (- π; π).

Solution. This equation is not homogeneous, so we will carry out transformations. The number 2 contained on the right side of the equation is replaced by the product 2 1

Since, according to the basic trigonometric identity, sin 2 x + cos 2 x \u003d 1, then

2 ∙ 1= 2 ∙ (sin 2 x + cos 2 x) = opening the brackets we get: 2 sin 2 x + 2 cos 2 x.

2 ∙ 1= 2 ∙ (sin 2 x + cos 2 x) =2 sin 2 x + 2 cos 2 x

So the equation 3sin 2 2x - 2 sin2xcos2 x +3cos 2 2x= 2 will take the form:

3sin 2 2x - 2sin 2x cos2 x +3cos 2 2x = 2 sin 2 x + 2 cos 2 x.

3sin 2 2x - 2 sin 2x cos2 x +3cos 2 2x - 2 sin 2 x - 2 cos 2 x=0,

sin 2 2x - 2 sin 2x cos2 x + cos 2 2x =0.

We have obtained a homogeneous trigonometric equation of the second degree. Let's apply the term-by-term division by cos 2 2x:

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

Let us introduce a new variable z= tg2х.

We have z 2 - 2 z + 1 = 0. This is a quadratic equation. Noticing the abbreviated multiplication formula on the left side - the square of the difference (), we get (z - 1) 2 = 0, i.e. z = 1. Let's return to the reverse substitution:

We have reduced the equation to the form tg t \u003d a, where t \u003d 2x, a \u003d 1. And since this equation has a solution for any value but and these solutions look like

x \u003d arctg x a + πn, then the solution to our equation will be:

2x \u003d arctg1 + πn,

x \u003d + , (x is equal to the sum of pi times eight and pi en times two).

It remains for us to find such values of x that are contained in the interval

(- π; π), i.e. satisfy the double inequality - π x π. Because

x= + , then - π + π. Divide all parts of this inequality by π and multiply by 8, we get

move the unit to the right and to the left, changing the sign to minus one

divide by four we get

for convenience, in fractions, we select integer parts

This inequality is satisfied by the following integer n: -2, -1, 0, 1

The last detail on how to solve tasks C1 from the exam in mathematics - solution of homogeneous trigonometric equations. We will tell you how to solve them in this final lesson.

What are these equations? Let's write them down in general terms.

$$a\sin x + b\cos x = 0,$$

where `a` and `b` are some constants. This equation is called a homogeneous trigonometric equation of the first degree.

To solve such an equation, you need to divide it by `\cos x`. Then it will take the form

$$\newcommand(\tg)(\mathop(\mathrm(tg))) a \tg x + b = 0.$$

The answer to such an equation is easily written in terms of the arc tangent.

Note that `\cos x ≠0`. To verify this, we substitute zero instead of cosine in the equation and we get that the sine must also be equal to zero. However, they cannot be equal to zero at the same time, which means that the cosine is not zero.

Some tasks of this year's real exam were reduced to a homogeneous trigonometric equation. Follow the link to. Let's take a slightly simplified version of the problem.

$$\sin x + \cos x = 0.$$

Divide by `\cos x`.

$$\tg x + 1 = 0,$$

$$x = -\frac(\pi)(4)+\pi k.$$

I repeat, a similar task was on the exam :) of course, you still need to select the roots, but this should not cause any particular difficulties either.

Let's now move on to the next type of equation.

In general, it looks like this:

$$a\sin^2 x + b\sin x \cos x + c\cos^2 x =0,$$

where `a, b, c` are some constants.

Such equations are solved by dividing by `\cos^2 x` (which again is non-zero). Let's take a look at an example right away.

$$\sin^2 x - 2\sin x \, \cos x - 3\cos^2 x = 0.$$

Divide by `\cos^2 x`.

$$(\tg)^2 x - 2\tg x -3 =0.$$

Replace `t = \tg x`.

$$t^2 - 2t -3 = 0,$$

$$t_1 = 3, \t_2 = -1.$$

Reverse replacement

$$\tg x = 3, \text( or ) \tg x = -1,$$

$$x = \arctan(3)+\pi k, \text( or ) x= -\frac(\pi)(4)+ \pi k.$$

Answer received.

$$-\sin^2 x + \frac(2\sqrt(2))(3)\sin x \cos x - 3\cos^2 x = -2.$$

Everything would be fine, but this equation is not homogeneous - we are hindered by `-2` on the right side. What to do? Let's use the basic trigonometric identity and write `-2` with it.

$$-\sin^2 x + \frac(2\sqrt(2))(3)\sin x \cos x - 3\cos^2 x = -2(\sin^2 x + \cos^2 x ),$$

$$-\sin^2 x + \frac(2\sqrt(2))(3)\sin x \cos x - 3\cos^2 x + 2\sin^2 x + 2\cos^2 x = 0,$$

$$\sin^2 x + \frac(2\sqrt(2))(3)\sin x \cos x - \cos^2 x = 0.$$

Divide by `\cos^2 x`.

$$(\tg)^2 x + \frac(2\sqrt(2))(3) \tg x - 1 = 0,$$

Replacing `t= \tg x`.

$$t^2 + \frac(2\sqrt(2))(3) t - 1 = 0,$$

$$t_1 = \frac(\sqrt(3))(3),\ t_2 = -\sqrt(3).$$

Making the reverse substitution, we get:

$$\tg x = \frac(\sqrt(3))(3) \text( or ) \tg x = -\sqrt(3).$$

$$x =-\frac(\pi)(3) + \pi k,\ x = \frac(\pi)(6)+ \pi k.$$

This is the last example in this tutorial.

As usual, let me remind you: training is our everything. No matter how brilliant a person is, skills will not develop without training. In an exam, this is fraught with excitement, mistakes, wasted time (continue this list yourself). Be sure to get busy!

Last name first name

cognitive activity

Solving Equations

Independent
Job

Solve the equations:

`10^(\sin x) = 2^(\sin x) \cdot 5^(-\cos x)`. This is a task from the real Unified State Examination 2013. Nobody canceled the knowledge of the properties of the degrees, but if you forgot, peep;
`\sqrt(3) \sin x + \sin^2 \frac(x)(2) = \cos^2 \frac(x)(2)`. Useful formula from the seventh lesson.
`\sqrt(3) \sin 2x + 3 \cos 2x = 0`.

That's all. And as usual, in the end: ask questions in the comments, put likes, watch videos, learn how to solve the exam.

"The greatness of a man is in his ability to think."
Blaise Pascal.

Lesson Objectives:

1) Educational- to introduce students to homogeneous equations, to consider methods for solving them, to contribute to the formation of skills for solving previously studied types of trigonometric equations.

2) Educational- to develop the creative activity of students, their cognitive activity, logical thinking, memory, the ability to work in a problem situation, to achieve the ability to correctly, consistently, rationally express their thoughts, to expand the horizons of students, to raise the level of their mathematical culture.

3) Educational- to cultivate the desire for self-improvement, diligence, to form the ability to competently and accurately perform mathematical records, to cultivate activity, to promote interest in mathematics.

Lesson type: combined.

Equipment:

Punched cards for six students.
Cards for independent and individual work of students.
Stands "Solution of trigonometric equations", "Numerical unit circle".
Electrified tables on trigonometry.
Presentation for the lesson (Attachment 1).

During the classes

1. Organizational stage (2 minutes)

Mutual greeting; checking the readiness of students for the lesson (workplace, appearance); organization of attention.

The teacher tells the students the topic of the lesson (slide 2) and explains that the handout that is on the desks will be used during the lesson.

2. Repetition of theoretical material (15 minutes)

Tasks on punched cards(6 people) . Working time on punched cards - 10 min (Annex 2)

By solving tasks, students will learn where trigonometric calculations are applied. The following answers are obtained: triangulation (a technique that allows measuring distances to nearby stars in astronomy), acoustics, ultrasound, tomography, geodesy, cryptography.

(slide 5)

front poll.

What equations are called trigonometric?
What types of trigonometric equations do you know?
What equations are called the simplest trigonometric equations?
What equations are called quadratic trigonometric?
Formulate the definition of the arcsine of a.
Formulate the definition of the arc cosine of a.
Formulate the definition of the arc tangent of a.
Formulate the definition of the inverse tangent of a.

Game "Guess the cipher word"

Blaise Pascal once said that mathematics is such a serious science that one should not miss an opportunity to make it a little more entertaining. So I suggest you play. After solving the examples, determine the sequence of digits by which the encrypted word is composed. In Latin, this word means "sine". (slide 3)

2) arctan (-√3)

4) tg(arc cos(1/2))

5) tg (arc ctg √3)

Answer: "Bend"

The game "Scattered mathematician»

Tasks for oral work are projected on the screen:

Check the correctness of the solution of the equations.(the correct answer appears on the slide after the student's answer). (slide 4)

	Answers with errors	Right answers
	x = ± π/6+2πn	x = ± π/3+2πn
	x = π/3+πn	X = (-1) nπ/3+πn
tg x = π/4	x = 1 +πn	tg x \u003d 1, x \u003d π / 4 + πn
	x = ±π/6+ π n	x = ± π/6+2πn
	x \u003d (-1) n arcsin1 / 3 + 2πn	x \u003d (-1) n arcsin1 / 3 + pn
	x = ± π/6+2πn	x = ± 5π/6+2πn
cos x = π/3	x = ± 1/2 +2πn	cos x = 1/2, x = ± π/3+2πn

Checking homework.

The teacher establishes the correctness and awareness of homework by all students; identifies gaps in knowledge; improves the knowledge, skills and abilities of students in the field of solving the simplest trigonometric equations.

1 equation. The student comments on the solution of the equation, the lines of which appear on the slide in the order of the comment). (slide 6)

√3tg2x = 1;

tg2x=1/√3;

2х= arctg 1/√3 +πn, n ∈Z.

2x \u003d π / 6 + πn, n ∈Z.

x \u003d π / 12 + π/2 n, n ∈Z.

2 equation. Solution h written to students on the board.

2 sin 2 x + 3 cosx = 0.

3. Actualization of new knowledge (3 minutes)

Students, at the request of the teacher, recall ways to solve trigonometric equations. They choose those equations that they already know how to solve, name the method of solving the equation and the result . The answers appear on the slide. (slide 7) .

Introduction of a new variable:

No. 1. 2sin 2x - 7sinx + 3 = 0.

Let sinx = t, then:

2t 2 – 7t + 3 = 0.

Factorization:

№2. 3sinx cos4x – cos4x = 0;

cos4x(3sinx - 1) = 0;

cos4x = 0 or 3 sinx - 1 = 0; …

No. 3. 2 sinx - 3 cosx = 0,

No. 4. 3 sin 2 x - 4 sinx cosx + cos 2 x \u003d 0.

Teacher: You do not know how to solve the last two types of equations yet. Both are of the same type. They cannot be reduced to an equation for the sinx or cosx functions. Are called homogeneous trigonometric equations. But only the first is a homogeneous equation of the first degree, and the second is a homogeneous equation of the second degree. Today in the lesson you will get acquainted with such equations and learn how to solve them.

4. Explaining new material (25 minutes)

The teacher gives students the definitions of homogeneous trigonometric equations, introduces the ways to solve them.

Definition. An equation of the form a sinx + b cosx =0, where a ≠ 0, b ≠ 0 is called homogeneous trigonometric equation of the first degree.(slide 8)

An example of such an equation is Equation #3. Let us write out the general form of the equation and analyze it.

and sinx + b cosx = 0.

If cosx = 0, then sinx = 0.

– Could such a situation happen?

- Not. We have obtained a contradiction to the basic trigonometric identity.

Hence, cosx ≠ 0. Let's perform term-by-term division by cosx:

a tgx + b = 0

tgx = -b / a is the simplest trigonometric equation.

Output: Homogeneous trigonometric equations of the first degree are solved by dividing both sides of the equation by cosx (sinx).

For example: 2 sinx - 3 cosx = 0,

Because cosx ≠ 0, then

tgx = 3/2 ;

x = arctg (3/2) + πn, n ∈Z.

Definition. An equation of the form a sin 2 x + b sinx cosx + c cos 2 x = 0 , where a ≠ 0, b ≠ 0, c ≠ 0 is called trigonometric equation of the second degree. (slide 8)

An example of such an equation is Equation #4. Let us write out the general form of the equation and analyze it.

a sin 2 x + b sinx cosx + c cos 2 x = 0.

If cosx = 0, then sinx = 0.

Again we got a contradiction to the basic trigonometric identity.

Hence, cosx ≠ 0. Let's perform term-by-term division by cos 2 x:

and tg 2 x + b tgx + c = 0 is a quadratic equation.

Conclusion: Oh homogeneous trigonometric equations of the second degree are solved by dividing both sides of the equation by cos 2 x (sin 2 x).

For example: 3 sin 2 x - 4 sinx cosx + cos 2 x \u003d 0.

Because cos 2 x ≠ 0, then

3tg 2 x - 4 tgx + 1 = 0 (Invite the student to go to the blackboard and complete the equation on their own).

Replacement: tgx = y. 3y 2 - 4y + 1 = 0

D = 16 - 12 = 4

y 1 = 1 or y 2 = 1/3

tgx=1 or tgx=1/3

x = arctg (1/3) + πn, n ∈Z.

x = arctg1 + πn, n ∈Z.

x = π/4 + πn, n ∈Z.

5. Stage of checking students' understanding of new material (1 min.)

Choose the extra equation:

sinx=2cosx; 2sinx + cosx = 2;

√3sinx + cosx = 0; sin 2 x - 2 sinx cosx + 4cos 2 x \u003d 0;

4cosx + 5sinx = 0; √3sinx – cosx = 0.

(slide 9)

6. Consolidation of new material (24 min).

Students, together with those answering at the blackboard, solve equations for new material. Tasks are written on the slide in the form of a table. When solving the equation, the corresponding part of the picture on the slide opens. As a result of the execution of 4 equations, a portrait of a mathematician who had a significant impact on the development of trigonometry opens up before students. (students will recognize the portrait of Francois Vieta - the great mathematician who made a great contribution to trigonometry, discovered the property of the roots of the reduced quadratic equation and was engaged in cryptography) . (slide 10)

1) √3sinx + cosx = 0,

Because cosx ≠ 0, then

√3tgx + 1 = 0;

tgx = –1/√3;

х = arctg (–1/√3) + πn, n ∈Z.

x = –π/6 + πn, n ∈Z.

2) sin 2 x - 10 sinx cosx + 21cos 2 x \u003d 0.

Because cos 2 x ≠ 0, then tg 2 x – 10 tgx + 21 = 0

Replacement: tgx = y.

y 2 - 10 y + 21 = 0

y 1 = 7 or y 2 = 3

tgx=7 or tgx=3

x = arctg7 + πn, n ∈Z

x = arctg3 + πn, n ∈Z

3) sin 2 2x - 6 sin2x cos2x + 5cos 2 2x = 0.

Because cos 2 2x ≠ 0, then 3tg 2 2x – 6tg2x +5 = 0

Replacement: tg2x = y.

3y 2 - 6y + 5 = 0

D \u003d 36 - 20 \u003d 16

y 1 = 5 or y 2 = 1

tg2x=5 or tg2x=1

2x = arctg5 + πn, n ∈Z

x = 1/2 arctg5 + π/2 n, n ∈Z

2x = arctg1 + πn, n ∈Z

x = π/8 + π/2 n, n ∈Z

4) 6sin 2 x + 4 sin(π-x) cos(2π-x) = 1.

6sin 2 x + 4 sinx cosx = 1.

6sin 2 x + 4 sinx cosx - sin 2 x - cos 2 x \u003d 0.

5sin 2 x + 4 sinx cosx - cos 2 x \u003d 0.

Because cos 2 x ≠0, then 5tg 2 x + 4 tgx –1 = 0

Replacement: tg x = y.

5y 2 + 4y - 1 = 0

D=16+20=36

y 1 = 1/5 or y 2 = -1

tgx = 1/5 or tgx = -1

x = arctg1/5 + πn, n ∈Z

x = arctg(–1) + πn, n ∈Z

x = –π/4 + πn, n ∈Z

Extras (on the card):

Solve the equation and, choosing one option from the four proposed, guess the name of the mathematician who derived the reduction formulas:

2sin 2 x - 3 sinx cosx - 5cos 2 x = 0.

Answer options:

х = arctg2 + 2πn, n ∈Z х = –π/2 + πn, n ∈Z – P. Chebyshev

x = arctan 12.5 + 2πn, n ∈Z x = –3π/4 + πn, n ∈Z – Euclid

х = arctg 5 + πn, n ∈Z х = –π/3 + πn, n ∈Z – Sofia Kovalevskaya

x = arctg2.5 + πn, n ∈Z x = –π/4 + πn, n ∈Z – Leonard Euler

Correct Answer: Leonhard Euler.

7. Differentiated independent work (8 min.)

The great mathematician and philosopher more than 2500 years ago suggested a way to develop mental abilities. “Thinking begins with wonder,” he said. We have been repeatedly convinced of the correctness of these words today. After completing independent work on 2 options, you can show how you learned the material and find out the name of this mathematician. For independent work, use the handout that is on your desks. You can choose one of the three proposed equations yourself. But remember that by solving the equation corresponding to yellow, you can only get "3", solving the equation corresponding to green - "4", red - "5". (Annex 3)

Whatever level of difficulty the students choose, after the correct solution of the equation, the first option gets the word "ARIST", the second - "HOTEL". On the slide the word is obtained: "ARIST-HOTEL". (slide 11)

Leaflets with independent work are handed over for verification. (Annex 4)

8. Recording homework (1 min)

D/z: §7.17. Compose and solve 2 homogeneous equations of the first degree and 1 homogeneous equation of the second degree (using Vieta's theorem for compilation). (slide 12)

9. Summing up the lesson, grading (2 minutes)

The teacher once again draws attention to those types of equations and those theoretical facts that were remembered in the lesson, speaks of the need to learn them.

Students answer the questions:

What kind of trigonometric equations are we familiar with?
How are these equations solved?

The teacher notes the most successful work in the lesson of individual students, puts marks.