Question

$left { {{x+y= pi } atop {sin^{2}x+sin^{2}y=1 }} right.$

Answer 1

$x+y=pi$
$y=pi-x$
$sin^2 x+sin^2 y=1$
$sin^2 x+sin^2 (pi-x)=1$
$sin^2 x+sin^2 x=1$
$2sin^2 x=1$
$sin^2 x=frac{1}{2}$
$frac{1-cos(2x)}{2}=frac{1}{2}$
$cos(2x)=0$
$2x=frac{pi}{2}+pi*k$
$x=frac{pi}{4}+frac{pi*k}{2}$
k є Z
$y=pi-x=frac{3pi}{4}+frac{pi*k}{2}$