#### Answer

$\{0^{o}, 30^{o}, 150^{o},180^{o}\}$

#### Work Step by Step

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 = $\{0^{o}, 30^{o}, 150^{o},180^{o}\}$