## Precalculus (6th Edition)

Solution set: $\{30^{o},210^{o},240^{o},300^{o}\}$
Using the zero product principle, either 1. $\quad \cot\theta-\sqrt{3}=0\quad$or 2. $\quad 2\sin+\sqrt{3=0.}$ 1. $\quad \cot\theta=\sqrt{3}$ $\theta=30^{o}\quad$or $\theta=30^{0}+180^{o}=210^{o}$ $2.\quad 2\sin\theta=-\sqrt{3}\quad/\div 2$ $\displaystyle \sin\theta=-\frac{\sqrt{3}}{2}$ ... use the unit circle, y coordinate = $-\displaystyle \frac{\sqrt{3}}{2}$... $\theta=240^{o} \quad$ or $\theta=300^{o}$ Solution set: $\{30^{o},210^{o},240^{o},300^{o}\}$