Centered at a point A.
α
- angle expressed in radians.
Definition
Sine (sin α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the opposite leg |BC| to the length of the hypotenuse |AC|.
Cosine (cos α) is a trigonometric function depending on the angle α between the hypotenuse and the leg of a right triangle, equal to the ratio of the length of the adjacent leg |AB| to the length of the hypotenuse |AC|.
Accepted notations
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Graph of the sine function, y = sin x
Graph of the cosine function, y = cos x
Properties of sine and cosine
Periodicity
Functions y = sin x and y = cos x periodic with period 2π.
Parity
The sine function is odd. The cosine function is even.
Domain of definition and values, extrema, increase, decrease
The sine and cosine functions are continuous in their domain of definition, that is, for all x (see proof of continuity). Their main properties are presented in the table (n - integer).
| y= sin x | y= cos x | |
| Scope and continuity | - ∞ < x < + ∞ | - ∞ < x < + ∞ |
| Range of values | -1 ≤ y ≤ 1 | -1 ≤ y ≤ 1 |
| Increasing | ||
| Descending | ||
| Maxima, y = 1 | ||
| Minima, y = - 1 | ||
| Zeros, y = 0 | ||
| Intercept points with the ordinate axis, x = 0 | y= 0 | y= 1 |
Basic formulas
Sum of squares of sine and cosine
Formulas for sine and cosine from sum and difference
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Formulas for the product of sines and cosines
Sum and difference formulas
Expressing sine through cosine
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Expressing cosine through sine
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Expression through tangent
; .
When , we have:
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At :
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Table of sines and cosines, tangents and cotangents
This table shows the values of sines and cosines for certain values of the argument. 
Expressions through complex variables
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Euler's formula
Expressions through hyperbolic functions
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Derivatives
; . Deriving formulas > > >
Derivatives of nth order:
{ -∞ <
x < +∞ }
Secant, cosecant
Inverse functions
The inverse functions of sine and cosine are arcsine and arccosine, respectively.
Arcsine, arcsin
Arccosine, arccos
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
Zakharova Lyudmila Vladimirovna
MBOU "Secondary school No. 59" of Barnaul
mathematic teacher
[email protected]
№1 The simplest trigonometric equations
Target: 1. Derive formulas for solutions of the simplest trigonometric equations of the form sinx =a, cosx=a, tgx=a, ctgx=a;
2. Learn to solve simple trigonometric equations using formulas.
Equipment: 1) Tables with graphs of trigonometric functions y= sinx, y=cosx, y=tgx, y=ctgx; 2) Table of values of inverse trigonometric functions; 3) Summary table of formulas for solving simple trigonometric equations.
Lecture lesson plan:
1 .Derivation of formulas for the roots of the equation
a) sinx =a,
b) cosx= a,
c) tgx= a,
d) ctgx= A.
2 . Oral frontal work to consolidate the received formulas.
3 . Written work to consolidate the studied material
During the classes.In algebra, geometry, physics and other subjects, we are faced with a variety of problems, the solution of which involves solving equations. We have studied the properties of trigonometric functions, so it is natural to turn to equations in which the unknown is contained under the function sign
Definition: Equations of the form sinx = a , cosx= a , tgx= a , ctgx= A are called the simplest trigonometric equations.
It is very important to learn how to solve the simplest trigonometric equations, since all methods and techniques for solving any trigonometric equations consist in reducing them to the simplest.
Let's start by deriving formulas that “actively” work when solving trigonometric equations.
1.Equations of the form sinx = a.
Let's solve the equation sinx = a graphically. To do this, in one coordinate system we will construct graphs of the functions y=sinx and y= A.
1) If A> 1 and A sin x= A has no solutions, since the straight line and the sine wave do not have common points.
2) If -1a a crosses the sine wave infinitely many times. This means that the equation sinx= a has infinitely many solutions.
Since the period of sine is 2 
, then to solve the equation sinx= a it is enough to find all solutions on any segment of length 2.
Solving the equation on [-/2; /2] by definition of arcsine x= arcsin a, and on x=-arcsin a. Taking into account the periodicity of the function у=sinx, we obtain the following expressions
x = -arcsin a+2n, n Z.
Both series of solutions can be combined
X = (-1) n arcsin a+n, nZ.
In the following three cases, they prefer to use simpler relations rather than a general formula:
If A=-1, then sin x =-1, x=-/2+2n
If A=1, then sin x =1, x =/2+2n
If a= 0, then sin x =0. x = n,
Example: Solve an equation sinx =1/2.
Let's create formulas for solutions x=arcsin 1/2+ 2n
X= - arcsin a+2n
Let's calculate the value arcsin1/2. Let's substitute the found value into the solution formulas
x=5/6+2n
or according to the general formula
X= (-1) n arcsin 1/2+n,
X= (-1) n /6+n,
2. Equations of the form cosx= a.
Let's solve the equation cosx= a also graphically, by plotting the functions y= cosx and y= A.
1) If a 1, then the equation cosx= a has no solutions, since the graphs do not have common points.
2) If -1 a cosx= a has an infinite number of solutions.
We will find all the solutions cosx= a on an interval of length 2 since the period of the cosine is 2.
By the definition of arc cosine, the solution to the equation will be x= arcos a. Considering the parity of the cosine function, the solution to the equation on [-;0] will be x=-arcos a.
Thus, solving the equation cosx= a x= + arcos a+ 2 n,
In three cases, we will not use the general formula, but simpler relations:
If A=-1, then cosx =-1, x =-/2+2n
If A=1, then cosx =1, x = 2n,
If a=0, then cosx=0. x =/2+n
Example: Solve an equation cos x =1/2,
Let's create formulas for solutions x=arccos 1/2+ 2n
Let's calculate the value arccos1/2.
Let's substitute the found value into the solution formulas
X= + /3+ 2n, nZ.
Equations of the form tgx= a.
Since the period of the tangent is equal, then in order to find all solutions to the equation tgx= a, it is enough to find all solutions on any interval of length . By definition of arctangent, the solution to the equation on (-/2; /2) is arctan a. Taking into account the period of the function, all solutions to the equation can be written in the form
x= arctan a+ n, nZ.
Example: Solve the equation tan x = 3/3
Let's create a formula for solving x= arctan 3/3 +n, nZ.
Let's calculate the arctangent value arctan 3/3= /6, then
X=/6+ n, nZ.
Derivation of the formula for solving the equation With tgx= a can be provided to students.
Example.
Solve the equation ctg x = 1.
x = arcсtg 1 + n, nZ,
X = /4 + n, nZ.
As a result of the material studied, students can fill out the table:
"Solving trigonometric equations."
the equation
Exercises to consolidate the studied material.
(Oral) Which of the written equations can be solved using the formulas:
a) x= (-1) n arcsin a+n, nZ;
b) x= + arcos a+ 2n?
cos x = 2/2, tan x= 1, sin x = 1/3, cos x = 3/3, sin x = -1/2, cos x= 2/3, sin x = 3, cos x = 2 .
Which of the following equations have no solutions?
Solve the equations:
a) sin x = 0; e) sin x = 2/2; h) sin x = 2;
b) cos x = 2/2; e) cos x = -1/2; i) cos x = 1;
d) tan x = 3; g) cot x = -1; j) tan x = 1/ 3.
3. Solve the equations:
a) sin 3x = 0; e) 2cos x = 1;
b) cos x/2 =1/2; e) 3 tg 3x =1;
d) sin x/4 = 1; g) 2cos(2x+ /5) = 3.
When solving these equations, it is useful to write down the rules for solving equations of the form sin V x = a, And With sin V x = a, | a|1.
Sin V x = a, |a|1.
V x = (-1) n arcsin a+n, nZ,
x= (-1) n 1/ V arcsin a+n/ V, nZ.
Summing up the lesson:
Today in class we derived formulas for solving simple trigonometric equations.
We looked at examples of solving simple trigonometric equations.
We filled out the table that we will use to solve the equations.
Homework.
№2 Solving trigonometric equations
Target: Study methods for solving trigonometric equations: 1) reducible to quadratic; 2) reducible to homogeneous trigonometric equations.
To develop students' powers of observation when using various methods of solving trigonometric equations.
Frontal work with students.
What are the formulas for the roots of trigonometric equations? cos x= a, sin x= a, tgx = a, ctg x = a.
Solve the equations (orally):
cos x=-1, sin x=0, tgx =0, cos x=1, cos x=1.5, sin x=0.
Find the errors and think about the reasons for the errors.
cos x=1/2, x= +
/6+2k,k 
Z.
sin x= 3/2, x= /3+k, kZ.
tgx = /4, x=1+ k, kZ.
2. Studying new material.
This lesson will cover some of the most common methods for solving trigonometric equations.
Trigonometric equations reduced to quadratic.
This class can include equations that include one function (sine or cosine) or two functions of the same argument, but one of them is reduced to the second using basic trigonometric identities.
For example, if cosх enters the equation in even powers, then we replace it with 1-sin 2 x, if sin 2 x, then we replace it with 1-cos 2 x.
Example.
Solve equation: 8 sin 2 x - 6sin x -5 =0.
Solution: Let's denote sin x=t, then 8t 2 - 6t – 5=0,
D= 196,
T 1 = -1/2, t 2 = -5/4.
Let's perform the reverse substitution and solve the following equations.
X=(-1) k+1 /6+ k, kZ.
Since -5/4>1, the equation has no roots.
Answer: x=(-1) k+1 /6+ k, kZ.
Solving consolidation exercises.
Solve the equation:
1) 2sin 2 x+ 3cos x = 0;
2) 5sin 2 x+ 6cos x -6 = 0;
3) 2sin 2 x+ 3cos 2 x = -2sin x;
4) 3 tg 2 x +2 tgx-1=0.
Homogeneous trigonometric equations.
Definition: 1) Equation of the forma sinx + b cosx=0, (a=0, b=0) is called a homogeneous equation of the first degree with respect to sin x and cos x.
This equation is solved by dividing both sides by cosx 
0. The result is the equation atgx+ b=0.
2) Equation of the forma sin 2 x + b sinx cosx + c cos 2 x =0 is called a homogeneous equation of the second degree, where a, b, c are any numbers.
If a = 0, then we solve the equation by dividing both sides by cos 2 x 0. As a result, we obtain the equation atg 2 x+ btgx+с =0.
Comment: Equation of the forma sin mx + b cos mx=0 or
a sin 2 mx + b sin mx cos mx + c cos 2 mx =0 are also homogeneous. To solve them, both sides of the equation are divided by cos mx=0 or cos 2 mx=0
3) Various equations that are not initially homogeneous can be reduced to homogeneous equations. For example,sin 2 mx + b sin mx cos mx + c cos 2 mx = d, And a sinx + b cosx= d. To solve these equations, you need to multiply the right-hand side by "trigonometric unit" those. on sin 2 x + cos 2 x and perform mathematical transformations.
Exercises to consolidate the learned material:
1) 2sin x- 3cos x = 0; 5) 4 sin 2 x – sin2x =3;
2) sin 2x+ cos2x = 0; 6) 3 sin 2 x + sinx cosx =2 cos 2 x ;
3) sin x+ 3cos x = 0; 7) 3 sin 2 x- sinx cosx =2;
4) sin 2 x -3 sinx cosx +2 cos 2 x =0
3. Summing up the lesson. Homework.
In this lesson, depending on the preparedness of the group, you can consider solving equations of the form a sin mx +b cos mx=c, where a, b, c are not equal to zero at the same time.
Strengthening exercises:
1. 3sin x + cos x=2;
2. 3sin 2x + cos 2x= 2;
3. sin x/3 + cos x/3=1;
4. 12 sin x +5 cos x+13=0.
№ 3 Solving trigonometric equations
Target: 1) Study the method of solving trigonometric equations by factorization; learn to solve trigonometric equations using various trigonometric formulas;
2) Check: students’ knowledge of formulas for solving simple trigonometric equations; ability to solve simple trigonometric equations.
Lesson plan:
Checking homework.
Mathematical dictation.
Learning new material.
Independent work.
Summing up the lesson. Homework.
Progress of the lesson:
Checking homework (solutions to trigonometric equations are briefly written on the board).
Mathematical dictation.
IN 1
1. What equations are called the simplest trigonometric equations?
2. What is the name of the equation of the forma sinx + b cosx=0? Indicate a way to solve it.
3.Write down the formula for the roots of the equation tgx = a(ctg x= a).
4. Write down the formulas for the roots of equations of the form cosx= a, Where A=1, A=0, A=-1.
5. Write down the general formula for the roots of the equation sin x= a, | a|
6. How equations of the form are solveda cosx= b, | b|
AT 2
1. Write down the formulas for the roots of the equations cosx= a,| a|
2. Write down the general formula for the roots of the equation
= a, | a|
3. What are equations of the form called? sin x= a, tgx = a, sin x= a?
4.Write down the formulas for the roots of the equation sin x= a, If A=1, A=0, A=-1.
5. How equations of the form are solved sin a x= b, | b|
6. What equations are called homogeneous equations of the second degree? How are they resolved?
Learning new material.
Factorization method.
One of the most commonly used methods for solving trigonometric equations is the factorization method.
If the equation f(x) =0 can be represented as f 1 (x) f 2 (x) =0, then the problem is reduced to solving two equations f 1 (x) = 0, f 2 (x) = 0.
(With students it is useful to remember the rule “ The product of factors is equal to zero if at least one of the factors is equal to zero, and the others make sense»)
Consolidation of the studied material through solving equations of varying complexity.
(sin x-1/2)(sin x+1)=0; 2) (cosx- 2/2)(sin x+ 2/2)=0;(self)
3) sin 2 x+ sin x cosx=0; 4) sin 2 x- sin x =0;
5) sin 2x – cosx=0; 6) 4 cos 2 x -1 =0; (2 ways)
7) cosx+ cos3x=0; 8) sin 3x= sin 17x;
9) sin x+ sin 2x+ sin 3x=0; 10) cos3x cos5x
11) sin x cos5x = sin 9x cos3x sin 2x sin 2x
12) 3 cosx sin x+ cos 2 x=0(self)
13) 2 cos 2 x - sin (x- /2)+ tanx tan (x+/2)=0.
Independent work.
Option-1 Option-2
1) 6 sin 2 x+ 5sin x -1=0; 1) 3 cos 2 x+2 cosx -5=0;
2) sin 2x – cos2x=0; 2) 3 cos x/2 - sin x/2=0;
3) 5 sin 2 x+ sin x cosx -2 cos 2 x=2; 3) 4sin 2 x- sin x cosx +7cos 2 x=5;
4) sin x+sin5x=sin3x+sin7x; 4) sin x-sin 2x +sin 3x-sin 4x=0;
5) sin x+cosx=1. 5) sin x+cosx=2.
8. Summing up the lesson. Homework.
The simplest trigonometric equations are solved, as a rule, using formulas. Let me remind you that the simplest trigonometric equations are:
sinx = a
cosx = a
tgx = a
ctgx = a
x is the angle to be found,
a is any number.
And here are the formulas with which you can immediately write down the solutions to these simplest equations.
For sine:
For cosine:
x = ± arccos a + 2π n, n ∈ Z
For tangent:
x = arctan a + π n, n ∈ Z
For cotangent:
x = arcctg a + π n, n ∈ Z
Actually, this is the theoretical part of solving the simplest trigonometric equations. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off the charts. Especially if the example deviates slightly from the template. Why?
Yes, because a lot of people write down these letters, without understanding their meaning at all! He writes down with caution, lest something happen...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)
Let's figure it out?
One angle will be equal to arccos a, second: -arccos a.
And it will always work out this way. For any A.
If you don’t believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number A to something negative. Anyway, we got one corner arccos a, second: -arccos a.
Therefore, the answer can always be written as two series of roots:
x 1 = arccos a + 2π n, n ∈ Z
x 2 = - arccos a + 2π n, n ∈ Z
Let's combine these two series into one:
x= ± arccos a + 2π n, n ∈ Z
And that's all. We have obtained a general formula for solving the simplest trigonometric equation with cosine.
If you understand that this is not some kind of superscientific wisdom, but just a shortened version of two series of answers, You will also be able to handle tasks “C”. With inequalities, with selecting roots from a given interval... There the answer with a plus/minus does not work. But if you treat the answer in a businesslike manner and break it down into two separate answers, everything will be resolved.) Actually, that’s why we’re looking into it. What, how and where.
In the simplest trigonometric equation
sinx = a
we also get two series of roots. Always. And these two series can also be recorded in one line. Only this line will be trickier:
x = (-1) n arcsin a + π n, n ∈ Z
But the essence remains the same. Mathematicians simply designed a formula to make one instead of two entries for series of roots. That's all!
Let's check the mathematicians? And you never know...)
In the previous lesson, the solution (without any formulas) of a trigonometric equation with sine was discussed in detail:
The answer resulted in two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
If we solve the same equation using the formula, we get the answer:
x = (-1) n arcsin 0.5 + π n, n ∈ Z
Actually, this is an unfinished answer.) The student must know that arcsin 0.5 = π /6. The complete answer would be:
x = (-1) n π /6+ π n, n ∈ Z
This raises an interesting question. Reply via x 1; x 2 (this is the correct answer!) and through lonely X (and this is the correct answer!) - are they the same thing or not? We'll find out now.)
We substitute in the answer with x 1 values n =0; 1; 2; etc., we count, we get a series of roots:
x 1 = π/6; 13π/6; 25π/6 and so on.
With the same substitution in response with x 2 , we get:
x 2 = 5π/6; 17π/6; 29π/6 and so on.
Now let's substitute the values n (0; 1; 2; 3; 4...) into the general formula for single X . That is, we raise minus one to the zero power, then to the first, second, etc. Well, of course, we substitute 0 into the second term; 1; 2 3; 4, etc. And we count. We get the series:
x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.
That's all you can see.) The general formula gives us exactly the same results as are the two answers separately. Just everything at once, in order. The mathematicians were not fooled.)
Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we won’t.) They are already simple.
I wrote out all this substitution and checking specifically. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a short summary of the answers. For this brevity, we had to insert plus/minus into the cosine solution and (-1) n into the sine solution.
These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these insertions can easily unsettle a person.
So what should I do? Yes, either write the answer in two series, or solve the equation/inequality using the trigonometric circle. Then these insertions disappear and life becomes easier.)
We can summarize.
To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing down the solution to an equation. For example, you need to solve the equations:
sinx = 0.3
Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z
cosx = 0.2
No problem: x = ± arccos 0.2 + 2π n, n ∈ Z
tgx = 1.2
Easily: x = arctan 1,2 + π n, n ∈ Z
ctgx = 3.7
One left: x= arcctg3,7 + π n, n ∈ Z
cos x = 1.8
If you, shining with knowledge, instantly write the answer:
x= ± arccos 1.8 + 2π n, n ∈ Z
then you are already shining, this... that... from a puddle.) Correct answer: there are no solutions. Don't understand why? Read what arc cosine is. In addition, if on the right side of the original equation there are tabular values of sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 and so on. - the answer through the arches will be unfinished. Arches must be converted to radians.
And if you come across inequality, like
then the answer is:
x πn, n ∈ Z
there is rare nonsense, yes...) Here you need to solve using the trigonometric circle. What we will do in the corresponding topic.
For those who heroically read to these lines. I simply cannot help but appreciate your titanic efforts. Bonus for you.)
Bonus:
When writing down formulas in an alarming combat situation, even seasoned nerds often get confused about where πn, And where 2π n. Here's a simple trick for you. In everyone formulas worth πn. Except for the only formula with arc cosine. It stands there 2πn. Two peen. Keyword - two. In this same formula there are two sign at the beginning. Plus and minus. Here and there - two.
So if you wrote two sign before the arc cosine, it’s easier to remember what will happen at the end two peen. And it also happens the other way around. The person will miss the sign ± , gets to the end, writes correctly two Pien, and he’ll come to his senses. There's something ahead two sign! The person will return to the beginning and correct the mistake! Like this.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
cos equation X = A
Each root of the equation
cos X = A (1)
can be considered as the abscissa of some intersection point of the sinusoid y = cosX with a straight line y =A , and, conversely, the abscissa of each such intersection point is one of the roots of equation (1). Thus, the set of all roots of equation (1) coincides with the set of abscissas of all intersection points of the cosine wave y = cosX with a straight line y = A .
If | A| >1 , then the cosine y = cosX does not intersect with a line y = A .
In this case, equation (1) has no roots.
At |A| < 1 there are infinitely many points of intersection.
for a > 0
for a< 0.
We will divide all these intersection points into two groups:
A -2 , A - 1 , A 1 , A 2 , ... ,
B -2 , B - 1 , B 1 , B 2 , ... ,
Dot A has an abscissa arccos A , and all other points of the first group are separated from it at distances that are multiples of 2 π
arccos a+ 2k π . (2)
Dot IN, as can be easily understood from the figures, has an abscissa - arccosA , and all other points of the second group are removed from it at distances that are multiples of 2 π . Therefore their abscissas are expressed as
arccos A+ 2nπ . (3)
Thus, equation (1) has two groups of roots defined by formulas (2) and (3). But these two formulas can obviously be written as one formula
X = ± arccos a+ 2m π , (4)
Where m runs through all integers (m = 0, ±1, ±2, ±3, ...).
The reasoning that we carried out in deriving this formula is correct only if
| a| =/= 1. However, formally the relation (4)
determines all the roots of the equation cosx=a
and at | A| =1. (Prove it!) Therefore we can say that the formula (4)
gives all roots of equation (1) for any values A
, If only |A|
<
1
.
But still in three special cases ( A = 0, A = -1, A= +1) we recommend not to use the formula (4) , but use other relations. It is useful to remember that the roots of the equation cos X = 0 are given by the formula
X = π / 2 +n π ; (5)
roots of the equation cos X = -1 are given by the formula
X = π + 2m π ; (6)
and finally, the roots of the equation cos X = 1 are given by the formula
X = 2m π ; (7)
In conclusion, we note that the formulas (4) , (5), (6) and (7) are correct only under the assumption that the desired angle X expressed in radians. If it is expressed in degrees, then these formulas need to be naturally changed. So, the formula (4) should be replaced by the formula
X = ± arccos a+ 360° n,
formula (5) formula
X = 90° + 180° n etc.
Examples:
\(2\sin(x) = \sqrt(3)\)
tg\((3x)=-\) \(\frac(1)(\sqrt(3))\)
\(4\cos^2x+4\sinx-1=0\)
\(\cos4x+3\cos2x=1\)
How to solve trigonometric equations:
Any trigonometric equation should be reduced to one of the following types:
\(\sint=a\), \(\cost=a\), tg\(t=a\), ctg\(t=a\)
where \(t\) is an expression with an x, \(a\) is a number. Such trigonometric equations are called the simplest. They can be easily solved using () or special formulas:
See infographics on solving simple trigonometric equations here:, and.
Example . Solve the trigonometric equation \(\sinx=-\)\(\frac(1)(2)\).Solution:
Answer: \(\left[ \begin(gathered)x=-\frac(π)(6)+2πk, \\ x=-\frac(5π)(6)+2πn, \end(gathered)\right.\) \(k,n∈Z\)
What each symbol means in the formula for the roots of trigonometric equations, see.
Attention! The equations \(\sinx=a\) and \(\cosx=a\) have no solutions if \(a ϵ (-∞;-1)∪(1;∞)\). Because sine and cosine for any x are greater than or equal to \(-1\) and less than or equal to \(1\):
\(-1≤\sin x≤1\) \(-1≤\cosx≤1\)
Example
. Solve the equation \(\cosx=-1,1\).
Solution:
\(-1,1<-1\), а значение косинуса не может быть меньше \(-1\). Значит у уравнения нет решения.
Answer
: no solutions.
Example . Solve the trigonometric equation tg\(x=1\).
Solution:
|
|
Let's solve the equation using the number circle. For this: |
Example
. Solve the trigonometric equation \(\cos(3x+\frac(π)(4))=0\).
Solution:
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Let's use the number circle again. \(3x+\)\(\frac(π)(4)\) \(=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\) \(3x+\)\(\frac(π)(4)\) \(=\)\(\frac(π)(2)\) \(+2πk\) \(3x+\)\(\frac( π)(4)\) \(=-\)\(\frac(π)(2)\) \(+2πk\) 8) As usual, we will express \(x\) in equations. \(3x=-\)\(\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\) \(3x=-\)\ (\frac(π)(4)\) \(+\)\(\frac(π)(2)\) \(+2πk\) |
Reducing trigonometric equations to the simplest is a creative task; here you need to use both and special methods for solving equations:
- Method (the most popular in the Unified State Examination).
- Method.
- Method of auxiliary arguments.
Let's consider an example of solving the quadratic trigonometric equation
Example . Solve the trigonometric equation \(2\cos^2x-5\cosx+2=0\)Solution:
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\(2\cos^2x-5\cosx+2=0\) |
Let's make the replacement \(t=\cosx\). |
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Our equation has become typical. You can solve it using . |
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\(D=25-4 \cdot 2 \cdot 2=25-16=9\) |
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\(t_1=\)\(\frac(5-3)(4)\) \(=\)\(\frac(1)(2)\) ; \(t_2=\)\(\frac(5+3)(4)\) \(=2\) |
We make a reverse replacement. |
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\(\cosx=\)\(\frac(1)(2)\); \(\cosx=2\) |
We solve the first equation using the number circle. |
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Let's write down all the numbers lying on at these points. |
An example of solving a trigonometric equation with the study of ODZ:
Example (USE) . Solve the trigonometric equation \(=0\)|
\(\frac(2\cos^2x-\sin(2x))(ctg x)\)\(=0\) |
There is a fraction and there is a cotangent - that means we need to write it down. Let me remind you that a cotangent is actually a fraction: ctg\(x=\)\(\frac(\cosx)(\sinx)\) Therefore, the ODZ for ctg\(x\): \(\sinx≠0\). |
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ODZ: ctg\(x ≠0\); \(\sinx≠0\)
\(x≠±\)\(\frac(π)(2)\) \(+2πk\); \(x≠πn\); \(k,n∈Z\) |
Let us mark the “non-solutions” on the number circle. |
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\(\frac(2\cos^2x-\sin(2x))(ctg x)\)\(=0\) |
Let's get rid of the denominator in the equation by multiplying it by ctg\(x\). We can do this, since we wrote above that ctg\(x ≠0\). |
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\(2\cos^2x-\sin(2x)=0\) |
Let's apply the double angle formula for sine: \(\sin(2x)=2\sinx\cosx\). |
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\(2\cos^2x-2\sinx\cosx=0\) |
If your hands reach out to divide by the cosine, pull them back! You can divide by an expression with a variable if it is definitely not equal to zero (for example, these: \(x^2+1.5^x\)). Instead, let's take \(\cosx\) out of brackets. |
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\(\cosx (2\cosx-2\sinx)=0\) |
Let’s “split” the equation into two. |
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\(\cosx=0\); \(2\cosx-2\sinx=0\) |
Let's solve the first equation using the number circle. Let's divide the second equation by \(2\) and move \(\sinx\) to the right side. |
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\(x=±\)\(\frac(π)(2)\) \(+2πk\), \(k∈Z\). \(\cosx=\sinx\) |
The resulting roots are not included in the ODZ. Therefore, we will not write them down in response. |
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We use a circle again. |
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These roots are not excluded by ODZ, so you can write them in the answer. |
