#### Answer

$\{90^{o}, 270^{o}\}$

#### Work Step by Step

$\sin^{2}\theta\cos\theta=\cos\theta\qquad/-\cos\theta$
$\sin^{2}\theta\cos\theta-\cos\theta=0$
$\cos\theta(\sin^{2}\theta-1)=0$
... zero factor principle ...
1. $\quad\cos\theta=0$ or
2.$\quad \sin^{2}\theta-1=0$
$1.\quad\cos\theta=0$ for $\theta=90^{o}, 270^{o}$
$2.\quad \sin^{2}\theta=1\quad/\pm\sqrt{(...)}$
$\sin\theta=\pm 1$
$\theta=90^{o}, 270^{o}$
Solution set = $\{90^{o}, 270^{o}\}$