Answer
$\theta\in\{90^{o},210^{o}, 330^{o}\}$
Work Step by Step
$\sin 3\theta=-1$
If $0^{o} \leq \theta < 360^{o}$ then
$0^{o} \leq 3\theta < 1080^{0}$
So, $3\theta\in[0,1080^{0})$, 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
$270^{o}$, but also
$270^{o}+360^{o}=630^{o}$, and
$270^{o}+720^{o}=990^{0}$
So, $3\theta\in\{ 270^{o},630^{o},990^{o}\}$,
and, since $(3\theta)\div 3=\theta,$
after dividing each corresponding angle with 3,
$\theta\in\{90^{o},210^{o}, 330^{o}\}$
