Hello, Dear friends! Today we will look at the task from part C. This is a system of two equations. The equations are quite peculiar. There are sine and cosine here, and there are also roots. The ability to solve quadratic and simple problems is required. In the presented task they detailed solutions are not presented, you should already be able to do this. Using the links provided, you can view the relevant theory and practical tasks.
The main difficulty in similar examples is that it is necessary to compare the obtained solutions with the found domain of definition; here one can easily make a mistake due to inattention.
The solution to the system is always a pair(s) of numbers x and y, written as (x;y).Be sure to check after receiving the answer.There are three ways presented to you, no, not ways, but three paths of reasoning that you can take. Personally, the third one is closest to me. Let's get started:
Solve the system of equations:
FIRST WAY!
Let's find the domain of definition of the equation. It is known that the radical expression has a non-negative meaning:
Consider the first equation:
1. It is equal to zero at x = 2 or at x = 4, but 4 radians does not belong to the definition of expression (3).
*An angle of 4 radians (229.188 0) lies in the third quarter, in which the sine value is negative. That's why
All that remains is the root x = 2.
Consider the second equation for x = 2.
At this value of x, the expression 2 – y – y 2 must be equal to zero, since
Let's solve 2 – y – y 2 = 0, we get y = – 2 or y = 1.
Note that for y = – 2 the root of cos y has no solution.
*An angle of –2 radians (– 114.549 0) lies in the third quarter, and in it the cosine value is negative.
Therefore, only y = 1 remains.
Thus, the solution to the system will be the pair (2;1).
2. The first equation is also equal to zero at cos y = 0, that is, at
But taking into account the found domain of definition (2), we obtain:
Consider the second equation for this y.
The expression 2 – y – y 2 with y = – Pi/2 is not equal to zero, which means that in order for it to have a solution the following condition must be met:
We decide:
Taking into account the found domain of definition (1), we obtain that
Thus, the solution to the system is one more pair:
SECOND WAY!
Let's find the domain of definition for the expression:
It is known that the expression under the root has a non-negative meaning.
Solving the inequality 6x – x 2 + 8 ≥ 0, we get 2 ≤ x ≤ 4 (2 and 4 are radians).
Consider Case 1:
Let x = 2 or x = 4.
If x = 4, then sin x< 0. Если х = 2, то sin x > 0.
Considering that sin x ≠ 0, it turns out that in this case in the second equation of the system 2 – y – y 2 = 0.
Solving the equation we find that y = – 2 or y = 1.
Analyzing the obtained values, we can say that x = 4 and y = – 2 are not roots, since we get sin x< 0 и cos y < 0 соответственно, а выражение стоящее под корнем должно быть ≥ 0 (то есть числом неотрицательным).
It can be seen that x = 2 and y = 1 are included in the domain of definition.
Thus, the solution is the pair (2;1).
Let's consider Case 2:
Let now 2< х < 4, тогда 6х – х 2 + 8 > 0. Based on this, we can conclude that in the first equation cos y should be equal to zero.
Solving the equation, we get:
In the second equation, when finding the domain of definition of the expression:
We get:
2 – y – y 2 ≥ 0
– 2 ≤ y ≤ 1
Of all the solutions to the equation cos y = 0, this condition is satisfied only by:
At given value y, expression 2 – y – y 2 ≠ 0. Therefore, in the second equation sin x will be equal to zero, we get:
Of all the solutions to this equation, the interval 2< х < 4 принадлежит только
This means that the solution to the system will be another couple:
*We didn’t find the domain of definition for all expressions in the system at once, we looked at the expression from the first equation (2 cases) and then along the way we determined the correspondence of the solutions found with established area definitions. In my opinion, it’s not very convenient, it turns out somehow confusing.
THIRD WAY!
It is similar to the first one, but there are differences. Also, the definition area for expressions is found first. Then the first and second equations are solved separately, and then the solution to the system is found.
Let's find the domain of definition. It is known that the radical expression has a non-negative meaning:
Solving the inequality 6x – x 2 + 8 ≥ 0 we get 2 ≤ x ≤ 4 (1).
Values 2 and 4 are radians, 1 radian as we know ≈ 57.297 0
In degrees we can approximately write 114.549 0 ≤ x ≤ 229.188 0.
Solving the inequality 2 – y – y 2 ≥ 0 we get – 2 ≤ y ≤ 1 (2).
In degrees we can write – 114.549 0 ≤ y ≤ 57.297 0 .
Deciding inequality sin x ≥ 0 we get that
Solving the inequality cos y ≥ 0 we get that
It is known that the product is equal to zero when one of the factors is equal to zero (and the others do not lose their meaning).
Consider the first equation:
Means
The solution to cos y = 0 is:
Solution 6x – x 2 + 8 = 0 are x = 2 and x = 4.
Consider the second equation:
Means
The solution to sin x = 0 is:
The solution to equation 2 – y – y 2 = 0 is y = – 2 or y = 1.
Now, taking into account the domain of definition, let’s analyze
obtained values:
Since 114.549 0 ≤ x ≤ 229.188 0, then this segment there is only one solution to the equation sin x = 0, this is x = Pi.
Since – 114.549 0 ≤ y ≤ 57.297 0, then this segment contains only one solution to the equation cos y = 0, this is
Consider the roots x = 2 and x = 4.
Right!
Thus, the solution to the system will be two pairs of numbers:
*Here, taking into account the found domain of definition, we excluded all obtained values that did not belong to it and then went through all the options for possible pairs. Next we checked which of them are the solution to the system.
I recommend immediately at the very beginning of solving equations, inequalities, and their systems, if there are roots, logarithms, trigonometric functions, be sure to find the domain of definition. There are, of course, examples where it is easier to solve immediately and then simply check the solution, but these are a relative minority.
That's all. Good luck to you!
The solution of trigonometric equations and systems of trigonometric equations is based on the solution of the simplest trigonometric equations.
Let us recall the basic formulas for solving the simplest trigonometric equations.
Solving equations of the form sin(x) = a.
When |a|< = 1 x = (-1)^k *arcsin(a) +π*k, где k принадлежит Z.
For |a|>1 there are no solutions.
Solving equations of the form cos(x) = a.
When |a|< = 1 x = ±arccos(a) +2*π*k, где k принадлежит Z.
For |a|>1 there are no solutions.
Solving equations of the form tg(x) = a.
x = arctan(a) + π*k, where k belongs to Z.
Solving equations of the form cotg(x) = a.
x = arcctg(a)+ π*k, where k belongs to Z.
Some common cases:
sin(x) =1; x = π/2 +2* π*k, where k belongs to Z.
sin(x) = 0; x = π*k, where k belongs to Z.
sin(x) = -1; x = - π/2 +2* π*k, where k belongs to Z.
cos(x) = 1; x = 2* π*k, where k belongs to Z.
cos(x) = 0; x= π/2 + π*k, where k belongs to Z.
cos(x) = -1; x = π+2* π*k, where k belongs to Z.
Let's look at a few examples:
Example 1. Solve the trigonometric equation 2*(sin(x))^2 + sin(x) -1 = 0.
Equations of this type are solved by reducing them to a quadratic equation by changing a variable.
Let y = sin(x). Then we get,
2*y^2 + y - 1 = 0.
We solve the resulting uvadratic equation using one of the known methods.
y1 = 1/2, y2 = -1.
Consequently, we obtain two simple trigonometric equations that can be solved using the formulas indicated above.
sin(x) = 1/2, x = ((-1)^k)*arcsin(1/2) + pi*k = ((-1)^k)*pi/6 + pi*k, for any whole k.
sin(x) = -1, x = - pi/2 +2* pi*n, where n belongs to Z.
Example 2. Solve the equation 6*(sin(x))^2 + 5*cos(x) – 2 = 0.
Using the basic trigonometric identity, we replace (sin(x))^2 with 1 - (cos(x))^2
We obtain a quadratic equation for cos(x):
6*(cos(x))^2 – 5*cos(x) - 4 = 0.
We introduce the replacement y=cos(x).
6*y^2 - 5*y - 4 = 0.
We solve the resulting quadratic equation y1 = -1/2, y2 = 1(1/3).
Since y = cos(x), and cosine cannot be more than one, we get one simple trigonometric equation.
x = ±2*pi/3+2*pi*k, for any integer k.
Example 3. tg(x) + 2*ctg(x) = 3.
Let's introduce the variable y = tan(x). Then 1/y = cot(x). We get
Multiply by y not equal to zero, we get a quadratic equation.
y^2 – 3*y + 2 = 0.
Let's solve it:
tg(x) = 2, x = arctan(2)+pi*k, for any integer k.
tg(x) = 1, x = arctan(1) + pi*k, pi/4 +pi*k, for any integer k.
Example 4. 3*(sin(x))^2 – 4*sin(x)*cos(x) + (cos(x))^2 = 0.
This equation can be reduced to a quadratic by dividing by either (cos(x))^2 or (sin(x))^2. When dividing by (cos(x)^2 we get
3*(tg(x))^2 – 4*tg(x) +1 = 0.
tg(x) = 1, x = pi/4+pi*n, for any integer n
tan(x) = 1/3, x = arctan(1/3) + pi*k, for any integer k.
Example 4. Solve a system of equations
( sin(x) = 2*sin(y)
From the bee-bread equation we express y,
Then we get, 2*sin(y) = 2*sin(x-5*pi/3) = 2*(sin(x)*cos(5*pi/3) - cos(x)*sin(5*pi /3)) = 2*(sin(x)*(1/2) –((√3)/2)*cos(x)) = sinx + √3*cos(x).
Lessons 54-55. Systems of trigonometric equations (optional)
09.07.2015 9098 895Target: consider the most typical systems trigonometric equations and methods for solving them.
I. Communicating the topic and purpose of the lessons
II. Repetition and consolidation of the material covered
1. Answers to questions about homework(analysis of unsolved problems).
2. Monitoring the assimilation of the material (independent work).
Option 1
Solve the inequality:
Option 2
Solve the inequality:
III. Learning new material
In exams, systems of trigonometric equations are much less common than trigonometric equations and inequalities. There is no clear classification of systems of trigonometric equations. Therefore, we will conditionally divide them into groups and consider ways to solve these problems.
1. The simplest systems of equations
These include systems in which either one of the equations is linear, or the equations of the system can be solved independently of each other.
Example 1
Let's solve the system of equations
Since the first equation is linear, we express the variable from itand substitute into the second equation:
We use the reduction formula and the main trigonometric identity. We get the equation or Let's introduce a new variable t = sin u. We have quadratic equation 3 t 2 - 7 t + 2 = 0, whose roots t 1 = 1/3 and t 2 = 2 (not suitable because sin y ≤ 1). Let's return to the old unknown and get the equation siny = 1/3, whose solution
Now it's easy to find the unknown:So, the system of equations has solutions where n ∈ Z.
Example 2
Let's solve the system of equations 
The equations of the system are independent. Therefore, we can write down the solutions to each equation. We get:
We add and subtract the equations of this system of linear equations term by term and find:
where 
Please note that due to the independence of the equations, when finding x - y and x + y, different integers must be specified n and k. If instead of k was also supplied n , then the solutions would look like:
In this case, an infinite number of solutions would be lost and, in addition, a connection between the variables would arise x and y: x = 3y (which is not the case in reality). For example, it is easy to check that this system has a solution x = 5π and y = n (in accordance with the formulas obtained), which when k = n impossible to find. So be careful.
2. Type systems 
Such systems are reduced to the simplest by adding and subtracting equations. In this case we obtain systems
or 
Let's note an obvious limitation: And The solution of such systems itself does not present any difficulties.
Example 3
Let's solve the system of equations 
Let us first transform the second equation of the system using the equality We get: 
Let's substitute the first equation into the numerator of this fraction:
and express 
Now we have a system of equations
Let's add and subtract these equations. We have: 
or
Let us write down the solutions to this simplest system:
Adding and subtracting these linear equations, we find:
3. Type systems
Such systems can be considered as simplest and solved accordingly. However, there is another way to solve it: convert the sum of the trigonometric functions into a product and use the remaining equation.
Example 4
Let's solve the system of equations 
First, we transform the first equation using the formula for the sum of the sines of angles. We get:
Using the second equation, we have:
where 
Let us write down the solutions to this equation:Taking into account the second equation of this system, we obtain a system of linear equations
From this system we find 
It is convenient to write such solutions in more rational form. For the upper signs we have:
for lower signs -
4. Type systems
First of all, it is necessary to obtain an equation containing only one unknown. To do this, for example, let us express from one equation sin y, from another - cos u. Let's square these ratios and add them up. Then we get a trigonometric equation containing the unknown x. Let's solve this equation. Then, using any equation of this system, we obtain an equation for finding the unknown y.
Example 5
Let's solve the system of equations 
Let us write the system in the form
Let us square each equation of the system and get:
Let's add up the equations of this system: or Using the basic trigonometric identity, we write the equation in the form or 
Solutions to this equation cos x = 1/2 (then 
) and cos x = 1/4 (from where 
), where n, k ∈ Z . Considering the connection between the unknowns cos y = 1 – 3 cos x, we get: for cos x = 1/2 cos y = -1/2; for cos x = 1/4 cos y = 1/4. It must be remembered that when solving a system of equations, squaring was carried out and this operation could lead to the appearance of extraneous roots. Therefore, it is necessary to take into account the first equation of this system, from which it follows that the quantities sin x and sin y must have the same sign.
Taking this into account, we obtain solutions to this system of equations
And 
where n, m, k, l ∈ Z . In this case, for unknown x and y, either upper or lower signs are simultaneously chosen.
In a special case
the system can be solved by converting the sum (or difference) of trigonometric functions into a product and then dividing the equations term by term.
Example 6
Let's solve the system of equations
In each equation, we transform the sum and difference of the functions into a product and divide each equation by 2. We get:
Since not a single factor on the left sides of the equations is equal to zero, we divide the equations term by term (for example, the second by the first). We get:
where 
Let's substitute the found valuefor example, in the first equation:
Let's take into account that 
Then 
where 
We obtained a system of linear equations
By adding and subtracting the equations of this system, we find
And 
where n, k ∈ Z.
5. Systems solved by replacing unknowns
If the system contains only two trigonometric functions or can be reduced to this form, then it is convenient to use the replacement of unknowns.
Example 7
Let's solve the system of equations
Since this system includes only two trigonometric functions, we introduce new variables a = tan x and b = sin u. We obtain a system of algebraic equations
From the first equation we express a = b + 3 and substitute into the second:
or 
The roots of this quadratic equation b 1 = 1 and b 2 = -4. The corresponding values are a1 = 4 and a2 = -1. Let's return to the old unknowns. We obtain two systems of simple trigonometric equations:
a) her decision 
where n, k ∈ Z.
b) has no solutions, because sin y ≥ -1.
Example 8
Let's solve the system of equations
Let us transform the second equation of the system so that it contains only the functions sin x and cos u. To do this, we use the reduction formulas. We get:
(where 
) And 
(Then 
). The second equation of the system has the form: or 
We obtained a system of trigonometric equations
Let's introduce new variables a = sin x and b = cos u. We have a symmetric system of equations 
only decision which a = b = 1/2. Let's go back to the old unknowns and get the simplest system trigonometric equations the solution of which 
where n, k ∈ Z.
6. Systems for which the features of the equations are important
Almost when solving any system of equations, one or another of its features is used. In particular, one of the most general techniques solutions of the system are identical transformations that make it possible to obtain an equation containing only one unknown. The choice of transformations, of course, is determined by the specifics of the system equations.
Example 9
Let's solve the system 
Let us pay attention to the left-hand sides of the equations, for example toUsing reduction formulas, we make it a function with argument π/4 + x. We get:
Then the system of equations looks like:
To eliminate the variable x, we multiply the equations term by term and get:
or 1 = sin 3 2у, whence sin 2у = 1. We find 
And 
It is convenient to consider separately the cases of even and odd values n. For even n (n = 2 k, where k ∈ Z) Then from the first equation of this system we obtain:
where m ∈ Z. For odd Then from the first equation we have:
So, this system has solutions
As in the case of equations, there are quite often systems of equations in which the limited nature of the sine and cosine functions plays a significant role.
Example 10
Let's solve the system of equations
First of all, we transform the first equation of the system:
or 
or or 
or 
Taking into account the limited nature of the sine function, we see that left side equation is not less than 2, and the right-hand side is not greater than 2. Therefore, such an equation is equivalent to the conditions sin 2 2x = 1 and sin 2 y = 1.
We write the second equation of the system in the form sin 2 y = 1 - cos 2 z or sin 2 y = sin 2 z, and then sin 2 z = 1. We obtained a system of simple trigonometric equationsUsing the formula for reducing the degree, we write the system in the form
or 
Then 
Of course, when solving other systems of trigonometric equations, it is also necessary to pay attention to the features of these equations.
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1 I. V. Yakovlev Materials on mathematics MathUs.ru Systems of trigonometric equations In this article we consider trigonometric systems of two equations with two unknowns. We will study methods for solving such systems and various special techniques immediately at specific examples. It may happen that one of the equations of the system contains trigonometric functions of the unknowns x and y, while the other equation is linear in x and y. In this case, we act in the obvious way: we express one of the unknowns from a linear equation and substitute it into another equation of the system. Problem 1. Solve the system: x + y =, sin x + sin y = 1. Solution. From the first equation we express y through x: and substitute it into the second equation: y = x, sin x + sin x) = 1 sin x = 1 sin x = 1. The result is the simplest trigonometric equation for x. We write its solutions in the form of two series: x 1 = 6 + n, x = n n Z). It remains to find the corresponding values of y: y 1 = x 1 = 5 6 n, y = x = 6 n. As always with a system of equations, the answer is given as a list of pairs x; y). 6 + n; 5) 5 6 n, 6 + n;) 6 n, n Z. Note that x and y are related to each other through the integer parameter n. Namely, if +n appears in the expression for x, then n automatically appears in the expression for y, and with the same n. This is a consequence of the “hard” relationship between x and y, given by the equation x + y =. Task. Solve the system: cos x + cos y = 1, x y =. Solution. Here it makes sense to first transform the first equation of the system: 1 + cos x cos y = 1 cos x + cos y = 1 cosx + y) cosx y) = 1. 1
2 Thus, our system is equivalent to the following system: cosx + y) cosx y) = 1, x y =. Substitute x y = into the first equation: cosx + y) cos = 1 cosx + y) = 1 x + y = n n Z). As a result, we arrive at the system: x + y = n, x y =. We add these equations, divide by and find x; subtract the second from the first equation, divide by and find y: x = + n, y = + n n Z). +n; + n), n Z. In a number of cases, the trigonometric system can be reduced to a system of algebraic equations by a suitable change of variables. Task. Solve the system: sin x + cos y = 1, sin x cos y = 1. Solution. The substitution u = sin x, v = cos y leads to an algebraic system for u and v: u + v = 1, u v = 1. You can easily solve this system yourself. The solution is unique: u = 1, v = 0. The reverse substitution leads to two simplest trigonometric equations: sin x = 1, cos y = 0, whence + k; + n), k, n Z. x = + k, y = + n k, n Z). Now the response record contains two integer parameters k and n. In contrast of previous tasks is that in this system there is no “rigid” connection between x and y, for example, in the form of a linear equation), therefore x and y are much more to a greater extent independent of each other.
3 V in this case It would be a mistake to use only one integer parameter n, writing the answer as + n;) + n. This would lead to loss infinite number 5 system solutions. For example, the solution would be lost ;) arising at k = 1 and n = 0. Problem 4. Solve the system: sin x + sin y = 1, cos x + cos y =. Solution. First we transform the second equation: 1 sin x + 1 sin y) = sin x + 4 sin y = 1. Now we make the replacement: u = sin x, v = sin y. We get the system: u + v = 1, u + 4v = 1. The solutions to this system are two pairs: u 1 = 0, v 1 = 1/ and u = /, v = 1/6. All that remains is to make the reverse substitution: sin x = 0, sin x = sin y = 1 or, sin y = 1 6, and write down the answer. k; 1) n 6 + n), 1) k arcsin + k; 1)n arcsin 16 + n), k, n Z. Problem 5. Solve the system: cos x + cos y = 1, sin x sin y = 4. Solution. Here, to obtain an algebraic system, you need to work even more. We write the first equation of our system in the form: In the second equation we have: cos x + y cos x y = 1. = sin x sin y = cosx y) cosx + y) = = cos x y 1 Thus, the original system is equivalent to the system: cos x + y cos x y = 1, cos x y cos x + y = 4. cos x + y) 1 = cos x y cos x + y.
4 We make the replacement u = cos x y, v = cos x + y and get an algebraic system: uv = 1, u v = 4. The solutions to this system are two pairs: u 1 = 1, v 1 = 1/ and u = 1, v = 1/. The first pair gives the system: x y = 1, = k, Hence cos x y cos x + y The second pair gives the system: cos x y cos x + y = 1 x + y x = ± + n + k), y = 1, = 1 = ± + n k, n Z). = ± + n k). x y = + k, x + y = ± + n k, n Z). Hence x = ± + n + k), y = ± + n k). ±) + n + k); ± + n k), ± + n + k); ±) + n k), k, n Z. However, it is not always possible to reduce a system of trigonometric equations to a system of algebraic equations. In some cases, it is necessary to use various special techniques. Sometimes it is possible to simplify a system by adding or subtracting equations. Problem 6. Solve the system: sin x cos y = 4, cos x sin y = 1 4. Solution. Adding and subtracting these equations, we get equivalent system: sinx + y) = 1, sinx y) = 1. And this system, in turn, is equivalent to a combination of two systems: x + y = + k, x + y = x y = + k, or 6 + n x y = n k, n Z). 4
5 Hence x = + k + n), x = + k + n), y = or + k n) y = + k n) k + n);)) 6 + k n), + k + n); + k n), k, n Z. 6 Sometimes you can come to a solution by multiplying equations by each other. Problem 7. Solve the system: tg x = sin y, ctg x = cos y. Solution. Let us recall that multiplying the equations of a system by each other means writing an equation of the form “the product of the left-hand sides is equal to the product of the right-hand sides.” The resulting equation will be a consequence of the original system, that is, all solutions of the original system satisfy the resulting equation). In this case, multiplying the equations of the system leads to the equation: 1 = sin y cos y = sin y, whence y = /4 + n n Z). It is inconvenient to substitute y in this form into the system; it is better to split it into two series: y 1 = 4 + n. Substitute y 1 into the first equation of the system: y = 4 + n. tan x = sin y 1 = 1 x 1 = 4 + k k Z). It is easy to see that substituting y 1 into the second equation of the system will lead to the same result. Now we substitute y: tan x = sin y = 1 x = 4 + k k Z). 4 + k;) 4 + n, 4) + k; 4 + n, k, n Z. Sometimes dividing the equations by each other leads to the result. Problem 8. Solve the system: cos x + cos y = 1, sin x + sin y =. Solution. Let's transform: cos x + y sin x + y cos x y cos x y = 1, =. 5
6 Let us temporarily introduce the following notation: α = x + y, β = x y. Then the resulting system will be rewritten in the form: cos α cos β = 1, sin α cos β =. It is clear that cos β 0. Then, dividing the second equation by the first, we arrive at the equation tg α =, which is a consequence of the system. We have: α = + n n Z), and again, for the purpose of further substitution into the system), it is convenient for us to divide the resulting set into two series: α 1 = + n, α = 4 + n. Substituting α 1 into any of the equations of the system leads to the equation: cos β = 1 β 1 = k k Z). Similarly, substituting α into any of the equations of the system gives the equation: cos β = 1 β = + k k Z). So, we have: that is, where α 1 = + n, β 1 = k or α = 4 + n, β = + k, x + y = + n, x + y = 4 x y or + n, = k x y = + k, x = + n + k), x = 7 + n + k), y = or + n k) y = + n k). + n + k);) 7 + n k), + n + k);) + n k), k, n Z. In some cases, the basic trigonometric identity comes to the rescue. Problem 9. Solve the system: sin x = 1 sin y, cos x = cos y. Solution. Let's square both sides of each equation: sin x = 1 sin y), cos x = cos y. 6
7 Let's add the resulting equations: = 1 sin y) + cos y = 1 sin y + sin y + cos y = sin y, whence sin y = 0 and y = n n Z). This is a consequence of the original system; that is, for any pair x; y), which is a solution to the system, the second number of this pair will have the form n with some integer n. We divide y into two series: y 1 = n, y = + n. We substitute y 1 into the original system: sin x = 1 sin y1 = 1, cos x = cos y1 = 1 The solution to this system is the series sin x = 1, cos x = 1. x 1 = 4 + k k Z). Please note that now it would not be enough to substitute y 1 into one of the equations of the system. Substituting y 1 into the first and second equations of the system leads to a system of two different equations relative to x.) Similarly, we substitute y into the original system: Hence sin x = 1 sin y = 1, cos x = cos y = 1 x = 4 + k k Z).)) 4 + k; n, + k; + n, k, n Z. 4 sin x = 1, cos x = 1. Sometimes, in the course of transformations, it is possible to obtain a simple relationship between unknowns and express from this relationship one unknown in terms of another. Problem 10. Solve the system: 5 cos x cos y =, sin x siny x) + cos y = 1. Solution. In the second equation of the system, we transform the double product of sines into the difference of cosines: cosx y) cos y + cos y = 1 cosx y) = 1 x y = n n Z). From here we express y in terms of x: y = x + n, 7
8 and substitute into the first equation of the system: 5 cos x cos x = 5 cos x cos x 1) = cos x 5 cos x + = 0. The rest is trivial. We get: cos x = 1, whence x = ± It remains to find y from the relation obtained above: + k k Z). y = ± + 4k + n. ± + k; ± + 4k + n), k, n Z. Of course, the considered problems do not cover the entire variety of systems of trigonometric equations. Any time difficult situation requires ingenuity, which is developed only by practice of solving various tasks. All answers assume that k, n Z. Problems 1. Solve the system: x + y =, cos x cos y = 1. b) x + y =, sin x sin y = 1. + n; n), + n; 4 n) ; b) n; n). Solve the system: x + y = 4, tg x tan y = 1 b) 6. x y = 5, sin x = sin y. arctan 1 + n; arctg 1 n), arctg 1 + n; arctg 1 n) ; b) + n; 6 + n). Solve the system: sin x + sin y = 1, x y = 4 b). x + y =, sin x sin y = n; 6 + n) ; b) 6 + n; 6 n) 8
9 4. Solve the system: sin x + cos y = 0, sin x + cos y = 1. b) sin x + cos y = 1, sin x cos y =. 1) k 6 + k; ± + n), 1) k k; ± + n) ; b) 1) k 4 + k; + n) 5. Solve the system: cos x + cos y = 1, tan x + tan y =, sin x sin y = b) 4. ctg x + ctg y = 9 5. ± + k; n) ; b) arctan 5 + k; arctan 1 + n), arctan 1 + k; arctan 5 + n) 6. Solve the system: sin x + cos y = 1, cos x cos y = 1. b) sin x + cos x = + sin y + cos y, sin x + sin y = 0. 1) k 6 + k; ± + n) ; b) 4 ± 4 + k; 5 4 ± 4 + n) 7. Solve the system: sin x + sin y =, cos x cos y = 1. 1) k 4 + k + n); 1)k 4 + k n)), 1) k k + n + 1); 1)k k n 1)) 8. Solve the system: sin x sin y = 1 4, tg x tan y =, cos x cos y = b) 4. sin x sin y = 4. ± 6 + k + n); ± 6 + k n)) ; b) ± + k + n); ± + k n)) 9. Solve the system: 4 sin x cos y = 1, tg x = tan y. b) sin x = cos x cos y, cos x = sin x sin y)k n k) ; 1) k 1 + n + k)) ; b)) 4 + k ; 4 + k + n 9
10 10. Solve the system: cos x = tan cos y = tan y +), 4 x +). 4k; n), 4 + k; 4 + n), + k; + n) 11. Solve the system:) tan 4 + x = cos y,) tan 4 x = sin y. k; 4 + n), + k; 4 + n) 1. Solve the system: sin x + sin y = 1, cos x cos y =. 6 + n + k); n k)), 6 + n + k); n k)) 1. Solve the system: tg x + tan y =, cos x cos y = n + k); 4 + n k)) 14. Solve the system: sin x = sin y, cos x = cos y. 6 + k; 4 + n), 6 + k; 4 + n), k; 4 + n), k; 4 + n) 15. Solve the system: 6 cos x + 4 cos y = 5, sin x + sin y = 0. arccos 4 + k; arccos n), arccos 4 + k; arccos n) 16. Solve the system: 4 tg x = tg y, sin x cosx y) = sin y. b) cot x + sin y = sin x, sin x sinx + y) = cos y. k; n); b)) 4 + k ; n, + k; + n) 10
11 17. “Fiztekh”, 010) Solve the system of equations 5 sin x cos y =, sin y + cos x =. 4 + k, 6 + n) ; k, n Z 18. Moscow State University, copy. for foreigners gr-n, 01) Solve the system of equations: 4 + cos x = 7 sin y, y x = y 4. + n; 6 + n), + n; n), + n; 6 n), + n; 5 6 n), n Z 19. MGU, VMK, 005) Find all solutions of the system sin equations x + y) = 1, xy = 9. xn, 4 + n) xn, where xn = 8 + n ± n) 6, n Z, n, 1, 0, 1 0. Moscow State University, geographical. f-t, 005) Solve the system of equations 1 sin x sin y =, 6 sin x + cos y =. 1) n n, k), k, n Z 1. Moscow State University, Faculty of State. control, 005) Solve the system of equations sin x sin 1 = 0, cos x cos 1 = n, n Z. MIPT, 199) Solve the system of equations 10 cos x = 7 cos x cos y, sin x = cos x sin y. arccos + n, 1)k arcsin 5); 6 + k arccos + n, 1)k+1 arcsin 5), 6 + k k, n Z 11
12 . MIPT, 199) Solve the system of equations tg x 4 ctg x = tg y, 4 sin x = sin x cos y. arctan 4 + n, arccos 4 + k) ; + arctan 4 + n, + arccos 4 + k), k, n Z 4. MIPT, 1996) Solve the system of equations sin x = sin y, cos y + cos x sin x = 4. ± 6 + n, 1)k k ) ; k, n Z 5. MIPT, 1996) Solve the system of equations sin x +) = sin y cos y, 4 sin y + sin x = 4 + sin x. 1) n 1 + n, 4 + 1)k 4 + k) ; k, n Z 6. MIPT, 1997) Solve the system of equations 9 cos x cos y 5 sin x sin y = 6, 7 cos x cos y sin x sin y = 4. ± n + k, ± 6 + n + k) ; k, n Z 1
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