Answer
$\{0^{o}, \ \ 90^{o}, \ \ 180^{0} , \ \ 270^{o}\}$
Work Step by Step
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: $\{0^{o}, \ \ 90^{o}, \ \ 180^{0} , \ \ 270^{o}\}$.
