Question

{ sinxcosy=-0,5{ siny*cosx = 0,5​

Answer 1

sinX·cosY=-0,5

cosX·sinY=0,5

Делим одно на другое, получаем:

tgX·ctgY=-1

tgX=-1÷ctgY

tgX=-tgY

X=-Y (или Y=-X)

Возвращаемся к одному из начальных уравнений и заменяем X на Y или Y на Х, пусть это будет первое, тогда:

sinX·cos(-Х)=-0,5 (косинус четна функция, т.е f(x)=f(-x))

sinX·cosX=-0,5

Воспользуемся формулой приведения  sinX·cosX=sin2X÷2

sinX·cosX=-0,5

sin2X÷2=-0,5

sin2X=-1

2X=-π÷2

X=-π÷4+2πn

Y=-X, т.е Y=π÷4+2πn

Answer 2

The given system of equations is:

sin(x)cos(y) = -0.5 siny*cos(x) = 0.5

Let's solve this system of equations:

From the second equation, we have:

siny*cos(x) = 0.5

Divide both sides by sin(y):

cos(x) = 0.5/siny

Now, substitute this value into the first equation:

sin(x) * (0.5/siny) = -0.5

Multiply both sides by sin(y):

sin(x) * (0.5/siny) * siny = -0.5 * siny

Simplify:

0.5 * sin(x) = -0.5 * siny

Divide both sides by 0.5:

sin(x) = -siny

Since sin(x) = -siny, we can substitute this back into the second equation:

-siny*cos(x) = 0.5

Rearrange the equation:

siny*cos(x) = -0.5

We can see that the second equation is the negative of the first equation. Therefore, the system of equations does not have a unique solution. There are infinitely many pairs of values for x and y that satisfy these equations.

In general, if you have a system of equations where one equation is the negative of the other, there will not be a unique solution.