The number of ordered pairs (x, y) of real numbers satisfying the...

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

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