## Precalculus (6th Edition)

$\{90^{o}, 270^{o}\}$
$\sin^{2}\theta\cos\theta=\cos\theta\qquad/-\cos\theta$ $\sin^{2}\theta\cos\theta-\cos\theta=0$ $\cos\theta(\sin^{2}\theta-1)=0$ ... 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 = $\{90^{o}, 270^{o}\}$