D = 4 - 4q.
x_1 =
![x_2 = \frac{-2 - \sqrt{4 - 4q} }{2} = \frac{-2 - 2\sqrt{1 - q} }{2} = -1 - \sqrt{1-q}](https://tex.z-dn.net/?f=x_2+%3D++%5Cfrac%7B-2+-++%5Csqrt%7B4+-+4q%7D+%7D%7B2%7D+%3D++%5Cfrac%7B-2+-++2%5Csqrt%7B1+-+q%7D+%7D%7B2%7D+%3D+-1+-++%5Csqrt%7B1-q%7D)
![x_2 - x_1 = -1 - \sqrt{1-q} - (-1 + \sqrt{1 - q} ) = -2 \sqrt{1 - q}](https://tex.z-dn.net/?f=x_2+-+x_1+%3D+-1+-+%5Csqrt%7B1-q%7D+-+%28-1+%2B+%5Csqrt%7B1+-+q%7D+%29+%3D+-2+%5Csqrt%7B1+-+q%7D+)
, т.е.
![x_2 - x_1 = -\sqrt{D}](https://tex.z-dn.net/?f=x_2+-+x_1+%3D+-%5Csqrt%7BD%7D+)
По обратной теореме Виета:
x₁ + x₂ = -2
x₂² - x₁² = (x₂ - x1)(x₁ + x₂)
![(-2 \sqrt{1-q} )*(-2) = 12](https://tex.z-dn.net/?f=%28-2+%5Csqrt%7B1-q%7D+%29%2A%28-2%29+%3D+12)
![\sqrt{1-q} =3](https://tex.z-dn.net/?f=+%5Csqrt%7B1-q%7D+%3D3)
![1-q = 9](https://tex.z-dn.net/?f=1-q+%3D+9)
![q = -8.](https://tex.z-dn.net/?f=q+%3D+-8.)
Ответ:
![x_1=2\pi n, n \in Z\\x_2=\frac{3\pi}{2}+2\pi k, k \in Z](https://tex.z-dn.net/?f=x_1%3D2%5Cpi%20n%2C%20n%20%5Cin%20Z%5C%5Cx_2%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%20k%2C%20k%20%5Cin%20Z)
Объяснение:
Преобразуем левую часть уравнения:
![cos(x)-sin(x)=\sqrt{1^2+1^2}*(\frac{1}{\sqrt{1^2+1^2}}cos(x)-\frac{1}{\sqrt{1^2+1^2}}sin(x))=\sqrt{2}(sin(\frac{\pi}{4})cos(x)-cos(\frac{\pi}{4})sin(x))=\sqrt{2}sin(\frac{\pi}{4}-x)](https://tex.z-dn.net/?f=cos%28x%29-sin%28x%29%3D%5Csqrt%7B1%5E2%2B1%5E2%7D%2A%28%5Cfrac%7B1%7D%7B%5Csqrt%7B1%5E2%2B1%5E2%7D%7Dcos%28x%29-%5Cfrac%7B1%7D%7B%5Csqrt%7B1%5E2%2B1%5E2%7D%7Dsin%28x%29%29%3D%5Csqrt%7B2%7D%28sin%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29cos%28x%29-cos%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29sin%28x%29%29%3D%5Csqrt%7B2%7Dsin%28%5Cfrac%7B%5Cpi%7D%7B4%7D-x%29)
Отсюда получим уравнение:
![\sqrt{2}sin(\frac{\pi}{4}-x)=1\\sin(\frac{\pi}{4}-x)=\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7Dsin%28%5Cfrac%7B%5Cpi%7D%7B4%7D-x%29%3D1%5C%5Csin%28%5Cfrac%7B%5Cpi%7D%7B4%7D-x%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
Найдем общее решение уравнения.
![sin(\frac{\pi}{4}-x)=\frac{\sqrt{2}}{2}\\sin(x-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\\x-\frac{\pi}{4}=(-1)^narcsin(-\frac{\sqrt{2}}{2})+\pi n\\x=\frac{\pi}{4}+(-1)^{n+1}\frac{\pi}{4}+\pi n, n \in Z](https://tex.z-dn.net/?f=sin%28%5Cfrac%7B%5Cpi%7D%7B4%7D-x%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5C%5Csin%28x-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%3D-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5C%5Cx-%5Cfrac%7B%5Cpi%7D%7B4%7D%3D%28-1%29%5Enarcsin%28-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%2B%5Cpi%20n%5C%5Cx%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%28-1%29%5E%7Bn%2B1%7D%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi%20n%2C%20n%20%5Cin%20Z)
Или же можно записать так:
![x_1=2\pi n, n \in Z\\x_2=\frac{3\pi}{2}+2\pi k, k \in Z](https://tex.z-dn.net/?f=x_1%3D2%5Cpi%20n%2C%20n%20%5Cin%20Z%5C%5Cx_2%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%20k%2C%20k%20%5Cin%20Z)
{2x+y=-1
{x+2y=2
x=2-2y
2(2-2y)+y=-1
4-4y+y=-1
-3y=-1-4
-3y=-5
y=<u>5 </u> = 1 ²/₃
3
x=2-2*<u> 5 </u>=2 - <u>10 </u>=<u> 6-10 </u>=<u> -4 </u>= -1 ¹/₃
3 3 3 3
Ответ: х= -1 ¹/₃
у= 1 ²/₃.