EntranceSolution of homogeneous trigonometric equations. Homogeneous trigonometric equations: general solution scheme
Lesson topic: "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
Organizing time
Greeting students, mobilizing attention.
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. .
Updating of basic knowledge.
Homework is checked and evaluated by an independent expert and consultants before the lesson and an evaluation sheet is filled out.
The teacher sums up the homework.
Teacher: 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.
Individual homework done in groups is checked. Defense of the presentation “Solutions of the simplest trigonometric equations”
(The work of the group is evaluated by an independent expert)
Motivation for learning.
Teacher: 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”.
Assimilation of new knowledge.
Teacher: The topic of the lesson is “Homogeneous trigonometric equations”.
Let's write the topic of the lesson in a notebook. Homogeneous trigonometric equations are of first and second degree.
Let us write down the definition of a homogeneous equation of the first degree. I use an example to show the solution of this type of equation, you make up an algorithm for solving a homogeneous trigonometric equation of the first degree.
Type equation a sinx + b cosx = 0 is called a homogeneous trigonometric equation of the first degree.
Consider the solution of the equation when the coefficients a and in different from 0.
Example: sinx + cosx = 0
R 
dividing both parts of the equation term by term by cosx, we obtain
Attention! Dividing by 0 is possible only if this expression does not turn into 0 anywhere. Let's analyze. If the cosine is 0, then it turns out that the sine will be 0, given that the coefficients are different from 0, but we know that the sine and cosine vanish in various points. Therefore, this operation can be performed when solving this type of equation.
Algorithm for solving a homogeneous trigonometric equation of the first degree: dividing both parts of the equation by cosx, cosx 0
Type equation a sin mx +b cos mx = 0 they also call a homogeneous trigonometric equation of the first degree and also solve the division of both parts of the equation by the cosine mx.
Type equation a sin 2 x +b sinx cox +c cos2x = 0 is called a homogeneous trigonometric equation of the second degree.
Example : sin 2 x + 2sinx cosx - 3cos 2 x=0
The coefficient a is different from 0 and therefore, as in the previous equation, cosx is not equal to 0, and therefore you can use the method of dividing both parts of the equation by cos 2 x.
We get tg 2 x + 2tgx - 3 = 0
We solve by introducing a new variable let tgx = a, then we get the equation
a 2 + 2a - 3 = 0
D \u003d 4 - 4 (-3) \u003d 16
a 1 = 1 a 2 = -3
Back to replacement
Answer:
If the coefficient a \u003d 0, then the equation will take the form 2sinx cosx - 3cos2x \u003d 0, we solve it by taking the common factor cosx out of brackets. If the coefficient c \u003d 0, then the equation will take the form sin2x + 2sinx cosx \u003d 0, we solve it by taking the common factor sinx out of brackets. Algorithm for solving a homogeneous trigonometric equation of the first degree:
See if the asin2 x term is in the equation.
If the term asin2 x is contained in the equation (i.e., a 0), then the equation is solved by dividing both sides of the equation by cos2x and then introducing a new variable.
If the term asin2 x is not contained in the equation (i.e., a = 0), then the equation is solved by the factorization method: cosx is taken out of brackets. Homogeneous equations of the form a sin2m x + b sin mx cos mx + c cos2mx = 0 are solved in the same way
The algorithm for solving homogeneous trigonometric equations is written in the textbook on page 102.
Physical education minute
Formation of skills for solving homogeneous trigonometric equations
Opening problem books page 53
1st and 2nd group decide No. 361-c
3rd and 4th group decide No. 363-v
Show the solution on the board, explain, supplement. An independent expert evaluates.
Solving examples from problem book No. 361-c
sinx - 3cosx = 0
we divide both sides of the equation by cosx 0, we get 
No. 363-v
sin2x + sinxcosx - 2cos2x = 0
we divide both sides of the equation by cos2x, we get tg2x + tgx – 2 = 0
solve by introducing a new variable
let tgx = a, then we get the equation
a2 + a - 2 = 0
D = 9
a1 = 1 a2 = -2
back to replacement
Solve equations.
2 cos - 2 = 0
2cos2x – 3cosx +1 = 0
3 sin2x + sinx cosx – 2 cos2x = 0
At the end of independent work, work and mutual verification change. Correct answers are displayed on the board.
Then they are handed over to an independent expert.
Do-It-Yourself Solution
Summing up the lesson.
What kind of trigonometric equations did we meet in the lesson?
Algorithm for solving trigonometric equations of the first and second degree.
Homework: § 20.3 read. No. 361 (d), 363 (b), increased difficulty additionally No. 380 (a).
Crossword.
If you enter the correct words, you get the name of one of the types of trigonometric equations.
The value of the variable that turns the equation into a true equality? (Root)
Angle unit? (Radian)
Numeric multiplier in product? (Coefficient)
Branch of mathematics that studies trigonometric functions? (Trigonometry)
What mathematical model is needed for the introduction trigonometric functions? (Circle)
Which of the trigonometric functions is even? (Cosine)
What is true equality called? (Identity)
Equality with a variable? (The equation)
Equations that have the same roots? (equivalent)
The set of roots of the equation ? (Decision)
Evaluation paper
№
n\n
Surname, name of the teacher
Presentation
cognitive activity
study
Solving Equations
Independent
Job
Homework - 12 points (3 equations 4 x 3 = 12 were given for homework)
Presentation - 1 point
Student activity - 1 answer - 1 point (4 points maximum)
Solving Equations 1 point
Independent work - 4 points
Group rating:
"5" - 22 points or more
“4” - 18 - 21 points
“3” - 12 - 17 points
Stop! Let's all the same try to understand this cumbersome formula.
In the first place should be the first variable in the degree with some coefficient. In our case, this
In our case it is. As we found out, it means that here the degree for the first variable converges. And the second variable in the first degree is in place. Coefficient.
We have it.
The first variable is exponential, and the second variable is squared, with a coefficient. This is the last term in the equation.
As you can see, our equation fits the definition in the form of a formula.
Let's look at the second (verbal) part of the definition.
We have two unknowns and. It converges here.
Let's consider all terms. In them, the sum of the degrees of the unknowns must be the same.
The sum of the powers is equal.
The sum of the powers is equal to (at and at).
The sum of the powers is equal.
As you can see, everything fits!
Now let's practice defining homogeneous equations.
Determine which of the equations are homogeneous:
Homogeneous equations - equations with numbers:
Let's consider the equation separately.
If we divide each term by expanding each term, we get
And this equation completely falls under the definition of homogeneous equations.
How to solve homogeneous equations?
Example 2
Let's divide the equation by.
According to our condition, y cannot be equal. Therefore, we can safely divide by
By substituting, we get a simple quadratic equation:
Since this is a reduced quadratic equation, we use the Vieta theorem:
Making the reverse substitution, we get the answer
Answer:
Example 3
Divide the equation by (by condition).
Answer:
Example 4
Find if.
Here you need not to divide, but to multiply. Multiply the whole equation by:
Let's make a replacement and solve the quadratic equation:
Making the reverse substitution, we get the answer:
Answer:
Solution of homogeneous trigonometric equations.
The solution of homogeneous trigonometric equations is no different from the solution methods described above. Only here, among other things, you need to know a little trigonometry. And be able to solve trigonometric equations (for this you can read the section).
Let's consider such equations on examples.
Example 5
Solve the equation.
We see a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.
Similar homogeneous equations are not difficult to solve, but before dividing the equations into, consider the case when
In this case, the equation will take the form: But sine and cosine cannot be equal at the same time, because according to the main trigonometric identity. Therefore, we can safely divide it into:
Since the equation is reduced, then according to the Vieta theorem:
Answer:
Example 6
Solve the equation.
As in the example, you need to divide the equation by. Consider the case when:
But the sine and cosine cannot be equal at the same time, because according to the basic trigonometric identity. So.
Let's make a substitution and solve the quadratic equation:
Let us make the reverse substitution and find and:
Answer:
Solution of homogeneous exponential equations.
Homogeneous equations are solved in the same way as those considered above. If you forgot how to decide exponential equations- see the relevant section ()!
Let's look at a few examples.
Example 7
Solve the Equation
Imagine how:
We see a typical homogeneous equation, with two variables and a sum of powers. Let's divide the equation into:
As you can see, after making the replacement, we get the reduced quadratic equation (in this case, there is no need to be afraid of dividing by zero - it is always strictly greater than zero):
According to Vieta's theorem:
Answer: .
Example 8
Solve the Equation
Imagine how:
Let's divide the equation into:
Let's make a replacement and solve the quadratic equation:
The root does not satisfy the condition. We make the reverse substitution and find:
Answer:
HOMOGENEOUS EQUATIONS. MIDDLE LEVEL
First, using an example of one problem, let me remind you what are homogeneous equations and what is the solution of homogeneous equations.
Solve the problem:
Find if.
Here you can notice a curious thing: if we divide each term by, we get:
That is, now there are no separate and, - now the desired value is the variable in the equation. And this is an ordinary quadratic equation, which is easy to solve using Vieta's theorem: the product of the roots is equal, and the sum is the numbers and.
Answer:
Equations of the form
called homogeneous. That is, this is an equation with two unknowns, in each term of which there is the same sum of the powers of these unknowns. For example, in the example above, this amount is equal to. The solution of homogeneous equations is carried out by dividing by one of the unknowns in this degree:
And the subsequent change of variables: . Thus, we obtain an equation of degree with one unknown:
Most often, we will encounter equations of the second degree (that is, quadratic), and we can solve them:
Note that dividing (and multiplying) the whole equation by a variable is possible only if we are convinced that this variable cannot be equal to zero! For example, if we are asked to find, we immediately understand that, since it is impossible to divide. In cases where this is not so obvious, it is necessary to separately check the case when this variable is equal to zero. For example:
Solve the equation.
Decision:
We see here a typical homogeneous equation: and are unknowns, and the sum of their powers in each term is equal.
But, before dividing by and getting the quadratic equation with respect, we must consider the case when. In this case, the equation will take the form: , hence, . But the sine and cosine cannot be equal to zero at the same time, because according to the basic trigonometric identity:. Therefore, we can safely divide it into:
I hope this solution is completely clear? If not, read the section. If it is not clear where it came from, you need to return even earlier - to the section.
Decide for yourself:
- Find if.
- Find if.
- Solve the equation.
Here I will briefly write directly the solution of homogeneous equations:
Solutions:
Answer: .
And here it is necessary not to divide, but to multiply:
Answer:
If you have not yet gone through trigonometric equations, you can skip this example.
Since here we need to divide by, we will first make sure that he does not zero:
And this is impossible.
Answer: .
HOMOGENEOUS EQUATIONS. BRIEFLY ABOUT THE MAIN
The solution of all homogeneous equations is reduced to division by one of the unknowns in the degree and further change of variables.
Algorithm:
You can order detailed solution your task!!!
An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tg x` or `ctg x`) is called a trigonometric equation, and we will consider their formulas 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's write the root formulas for each of them.
1. Equation `sin x=a`.
For `|a|>1` it has no solutions.
With `|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 the sine, there are no solutions among real numbers.
With `|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`
It 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 sinus: 
For cosine: 
For tangent and cotangent: 
Formulas for solving equations containing inverse trigonometric functions:
Methods for solving trigonometric equations
The solution of any trigonometric equation consists of two stages:
- using to convert it to the simplest;
- solve the resulting simple equation using the above formulas for the roots and tables.
Let's consider the main methods of solution using examples.
algebraic method.
In this method, the replacement of a variable and its substitution into equality is done.
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`.
Decision. Move to the left all terms of equality: `sin x+cos x-1=0`. Using , we transform and factorize the left 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`,
- `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
- `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 bring 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 split both parts by `cos x \ne 0` for the first case, and by `cos^2 x \ne 0` for the second. We get equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which must be solved using known methods.
Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.
Decision. 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, dividing its left and right parts 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`, as a result `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:
- `tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
- `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`.
Go to Half Corner
Example. Solve the equation: `11 sin x - 2 cos x = 10`.
Decision. Applying the double angle formulas, the result is: `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:
- `tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
- `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 an auxiliary angle
In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, we divide both parts 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 1 and their modulus is at most 1. Let's 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`.
Decision. Dividing both sides of the equation 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`.
Denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, 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, in the numerators and denominators of which there are trigonometric functions.
Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.
Decision. Multiply and divide the right side of the equation 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`
Given that the denominator cannot be zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.
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`.
- `sin x=0`, `x=\pi n`, `n \in Z`
- `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. The study begins in the 10th grade, there are always tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy for you!
However, you don’t even need to memorize them, the main thing is to understand the essence, and be able to deduce. It's not as difficult as it seems. See for yourself by watching the video.
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.
Download:
Preview:
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. Appendix 1.
Evaluation sheet No.
n\n | Last name first name | Homework | cognitive activity | Solving Equations | Independent Job | Grade |
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? (Decision)
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
Teacher: Sinitsina S.I.
MBOU secondary school No. 20 named after Milevsky N.I.
Topic: Homogeneous trigonometric equations (Grade 10)
Objectives: To 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 degree;
To consolidate the skills of solving all types of trigonometric equations through
development and improvement of skills to apply existing knowledge in a changed
situations, through the ability to draw conclusions and generalize
Education in students of accuracy, culture of behavior.
Type of lesson: a lesson in the formation of new knowledge.
Equipment: computer, multimedia projector, screen, whiteboard, presentation
During the classes
I. Organizational moment
Greeting students, mobilizing attention.
II. Updating of basic knowledge ( Homework is checked by consultants before the lesson. The teacher sums up the homework.)
Teacher: We continue to study 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.
oral work
- What equation do we call trigonometric?
- Name the algorithm for solving the equation cos t = a
- Name the algorithm for solving the equation sin t = a
III. Motivation for learning.
Teacher: 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. Having solved the crossword puzzle, the guys will read the word “homogeneous”.
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? (Equations)
9. Equations with the same roots? (equivalent)
10. Set of roots of the equation? (Decision)
IV. Explanation of the new topic
Teacher: The topic of the lesson is “Homogeneous trigonometric equations”.
Let's write the topic of the lesson in a notebook. Homogeneous trigonometric equations are of first and second degree.
Let us write down the definition of a homogeneous equation of the first degree. I use an example to show the solution of this type of equation, you make up an algorithm for solving a homogeneous trigonometric equation of the first degree.
Type equation and sinx + b cosx = 0 is called a homogeneous trigonometric equation of the first degree.
Consider the solution of the equation when the coefficients a and b are different from 0.
Example1: 2sinx - 3cosx = 0
Dividing both sides of the equation term by term by cosx, we obtain
2sinx/ cosx - 3cosx/ cosx = 0
2 tg x-3=0,tg x =3/2, x= arctg3/2 + πn, nє Z,
Attention! It is possible to divide by the same expression only if this expression does not turn to 0 anywhere. Let's analyze. If the cosine is 0, then in order for the whole expression to turn into 0, the sine must also be equal to 0 (we take into account that the coefficients are different from 0). But we know that sine and cosine vanish at different points. Therefore, such an operation can be performed when solving this type of equations.
Type equation and sin mx + b cos mx = 0 is also called a homogeneous trigonometric equation of the first degree and is also solved by dividing both parts of the equation by cos mx.
Type equation a sin 2 x + b sinx cosx + c cos 2 x = 0 is called a homogeneous trigonometric equation of the second degree.
Example2: sin 2 x – 3 sinx cosx +2 cos 2 x = 0
The coefficient a is different from 0 and therefore, as in the previous equation, cosx 0 and therefore you can use the method of dividing both parts of the equation by cos 2 x.
We get tg 2 x - 3 tgx +2 = 0
We solve by introducing a new variable let tgx = a, then we get the equation
a 2 -3 a +2 = 0 a 1 = 1 a 2 = 2
Back to replacement
tgx =1, x = ¼π+ πn, nє Z tgx = 2 , x = arctg 2 + πn, nє Z
Answer: x = ¼π + πn, nє Z, x = arctg 2 + πn, nє Z
If the coefficient a \u003d 0, then the equation will take the form -3sinx cosx + 2cos 2 x \u003d 0, we solve it by taking the common factor - cosx out of brackets: - cosx (3 sinx - 2cosx) \u003d 0,
cosx = 0 or 3sinx – 2cosx = 0. The second equation is a homogeneous equation of the first degree.
If the coefficient c \u003d 0, then the equation will take the form sin 2 x -3sinx cosx \u003d 0, we solve it by taking the common multiplier sinx out of brackets: sinx (sinx -3 cosx) \u003d 0,
sinx = 0 or sinx -3 cosx = 0. The second equation is a homogeneous equation of the first degree.
Algorithm for solving a homogeneous trigonometric equation of the second degree:
1. See if there is a term a sin 2 x in the equation.
2. If the term asin 2 x is contained in the equation (i.e., a 0), then the equation is solved by dividing
both parts of the equation on cos 2 x and the subsequent introduction of a new variable a = tgx
3. If the term asin 2 x is not contained in the equation (i.e., a = 0), then the equation is solved by the factorization method: cosx is taken out of brackets.
Homogeneous equations of the form a sin 2 mx + b sin mx cos mx + c cos 2 mx = 0 solved in the same way
V. Assimilation of new knowledge
Are these equations homogeneous?
- sin x = 2 cos x
- sin5x + cos5x = 0
- sin 3x - cos 3x = 2
- sin 2 8x – 5 sin8x cos8x +2 cos 2 8x =0
VI. Physical education
VII. Formation of skills for solving homogeneous trigonometric equations
We open the problem books p. 47 No. 18.10 (a), No. 18.11 (a, b), 18.12 (d)
VIII. Independent work ( students choose differentiated tasks according to two options)
1 option 2 option
1) sinx + 2cosx = 0. 1) sinx - 4cosx = 0.
2) sin 2 x + 2sinx cosx -3 cos 2 x = 0 2) sin 2 x - 4 sinx cosx +3 cos 2 x = 0
3) 2sin 2 2x – 5 sin2x cos2x +2 cos 2 2x = 0 3) 3sin 2 3x +10 sin3x cos3x +3 cos 2 3x = 0
Correct answers are displayed on the board.
IX. Summing up the lesson, grading
What kind of trigonometric equations did we meet in the lesson?
What equations do we call homogeneous?
Formulate algorithms for solving homogeneous trigonometric equations of the first and second degree.
X. Homework: Compose and solve 2 homogeneous equations of the first degree and 1 homogeneous equation of the second degree