## Precalculus (6th Edition)

$\{0^{o}, 30^{o}, 150^{o},180^{o}\}$
Factor out $\tan^{2}\theta$ on the LHS $\tan^{2}\theta (2\sin\theta-1)=0$ ... zero factor principle ... 1. $\quad\tan^{2}\theta=0$ or 2.$\quad 2\sin\theta-1=0$ 1. $\quad\tan^{2}\theta=0$ $\tan\theta=0\Rightarrow \theta=0^{o}, 180^{o}$ 2.$\quad 2\sin\theta-1=0\qquad/+1$ $2\sin\theta=1\qquad/\div 2$ $\displaystyle \sin\theta=\frac{1}{2}\qquad$ ... use the unit circle... $\theta=30^{o}, 150^{o}$ Solution set = $\{0^{o}, 30^{o}, 150^{o},180^{o}\}$