## Precalculus (6th Edition)

$\{0^{o}, \ \ 90^{o}, \ \ 180^{0} , \ \ 270^{o}\}$
As solved in exercise $4$, two angles (degree measures) with sine equal to 0 are $\{0^{o}, \ \ 180^{o}\}$ . Also, if $0 \leq \theta < 360^{o}\qquad/\times 2$ then $0 \leq 2\theta < 720^{o}.$ So either $2\theta= 0^{o}\quad$or$\quad 2\theta= 0^{o}+360^{o}$ or $2\theta= 180^{o}\quad$or$\quad 2\theta= 180^{o}+360$ $...$divide both sides by 2 in each equation $\theta= 0^{o}\quad$or$\quad \theta=180^{o}$ or $\theta=90^{o}\quad$or$\quad \theta= 90^{o}+180^{o}=270^{o}$ Solution set: $\{0^{o}, \ \ 90^{o}, \ \ 180^{0} , \ \ 270^{o}\}$.