$\sin x=\sin 2y$ and $\cos x=\sin y$

Squaring and adding, we get,
$1={\left(2\sin y\cos y\right)}^2+{\sin }^{2}y$
$1-{\sin }^{2}y=4{\sin }^{2}y{\cos }^{2}y$
${\cos }^{2}y\left(1-4{\sin }^{2}y\right)=0$
$\cos y=0$ or $\sin y=\pm \frac{1}{2}$
$y=\frac{\pi }{2}$ or $\frac{\pi }{6}$ or $\frac{5\pi }{6}$
$y=\frac{\pi }{2}\Rightarrow x=0$
$y=\frac{\pi }{6}\Rightarrow x=\frac{\pi }{3}$
$\Rightarrow$ two ordered pairs $\left(0,\frac{\pi }{2}\right) \& \left(\frac{\pi }{3},\frac{\pi }{6}\right)$