Answer
$\theta=\{120^o,150^o,300^o,330^o\}$
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 $360^o$
$2\theta=180^o+60^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+240^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o-60^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+300^o$
$\theta=120^o\;\;\;\;or\;\;\;\;\;\;\theta=300^o\;\;\;\;or\;\;\;\;\;\;\theta=150^o\;\;\;\;or\;\;\;\;\;\;\theta=330\;\;\;$
$\theta=\{120^o,150^o,300^o,330^o\}$
