Answer
$\theta=\{120^o+180^ok\;\;,\;\;150^o+180^ok\}$
Work Step by Step
$sin(2\theta)=\frac{-\sqrt{3}}{2}$
$2\theta=sin^{-1}(\frac{-\sqrt{3}}{2})$
We know $sin(\theta)$ is Negative in quardent $III$ and quardent $IV$
The period of the sine function is $360^o$
$2\theta=180^o+60^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o-60^o\;\;\;\; $
$2\theta=240^o+360^ok\;\;\;\;or\;\;\;\;\;\;2\theta=300^o+360^ok\;\;\;\; $
$\theta=120^o+180^ok\;\;\;\;or\;\;\;\;\;\;\theta=150^o+180^ok\;\;\;\; $
$\theta=\{120^o+180^ok\;\;,\;\;150^o+180^ok\},\;\;\;\;\;\;\;\;\;\;when \;\;\;K=\{0,1,2\}$
