#### Answer

$\theta\in\{0^{o},120^{o}, 240^{o}\}$

#### Work Step by Step

$\sin 3\theta=0$
If $0^{o} \leq \theta < 360^{o}$ then
$0^{o} \leq 3\theta < 1080^{0}$
So, $3\theta\in[0,1080^{o})$, meaning that
if $ 3\theta$ (a solution) is any angle from $[0^{o},360^{o})$, then
$(3\theta+360^{o})$ and $(3\theta+720^{o})$ are also solutions.
Using the unit circle,$ 3\theta$ can be
$0^{o}$, but also
$0^{o}+360^{o}=360^{o}$, and
$0^{o}+720^{o}=720^{0}$
So, $3\theta\in\{ 0^{o},360^{o},720^{o}\}$,
and, since $(3\theta)\div 3=\theta,$
after dividing each corresponding angle with 3,
$\theta\in\{0^{o},120^{o}, 240^{o}\}$