Question

Solve the following system.
$sinx+cosy=0,si{n}^{2}x+co{s}^{2}y=\frac{1}{2}$
$0<x<\pi ,0<y<\pi .$

Answer 1

$\sin x+\cos y=0⟹\cos y=-\sin x$

${\sin }^{2}x+{\cos }^{2}y=\frac{1}{2}⟹{\sin }^{2}x=\frac{1}{4}⟹\sin x=\frac{1}{2}⟹x=\frac{\pi }{6},\frac{5\pi }{6}$

$\cos y=-\frac{1}{2}⟹y=\frac{2\pi }{3}$