If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$  and $\sin 2x=\cos 2y$  then:

For the given equation  $0\le x,y\le 1$
Also  ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$
 $\Rightarrow {\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$
 $\Rightarrow {\sin }^{-1}x={\cos }^{-1}y={\sin }^{-1}\sqrt{1-y^2}$
 $\Rightarrow x=\sqrt{1-y^2}$
 $\Rightarrow x^2+y^2=1\mathrm{.....}\left(i\right)$
Again,  $\sin 2x=\cos 2y$
 $\Rightarrow \cos (\frac{\pi }{2}-2x)=\cos 2y$
$\Rightarrow \frac{\pi }{2}-2x=2n\pi +2y,$ where  $n\in I$
 $\Rightarrow x\pm y=\frac{\pi }{4}(\because x,y,\in [-1,1])\mathrm{.....}\left(ii\right)$
From eq. (i) and (ii), we get
  $x=\frac{\pi }{8}+\sqrt{\frac{1}{2}-\frac{{\pi }^2}{64}}$
And  $y=\sqrt{\frac{1}{2}-\frac{{\pi }^2}{64}}-\frac{\pi }{8}$