Вариант 3 , задание 3, заранее спасибо

Решите задачу:

$\left \{ {{x+y=\frac{\pi}{2}} \atop {sin x+sin y=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+sin y=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+sin (\frac{\pi}{2}-x)=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin x+cosx=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {\sqrt{2}(\frac{1}{\sqrt{2}}sin x+\frac{1}{\sqrt{2}}cosx)=-\sqrt{2}}} \right;\\$

$\left \{ {{y=\frac{\pi}{2}-x} \atop {\sqrt{2}(sin xcos\frac{\pi}{4}+cos xsin\frac{\pi}{4})}=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {\sqrt{2}sin (x+\frac{\pi}{4})}=-\sqrt{2}}} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {sin (x+\frac{\pi}{4})}=-1} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {x+\frac{\pi}{4}}=-\frac{\pi}{2}+2*\pi*k} \right;\\ \left \{ {{y=\frac{\pi}{2}-x} \atop {x=-\frac{3\pi}{4}+2*\pi*k} \right;\\$

$\left \{ {y=\frac{5\pi}{4}-2 *\pi*k} \atop {x=-\frac{3\pi}{4}+2*\pi*k} \right;\\ \left \{ {y=\frac{-3\pi}{4}-2 *\pi*k} \atop {x=-\frac{3\pi}{4}+2*\pi*k} \right;\\$