Answer
$\theta=\{6^o+36^ok\;\;,\;\;12^o+36^ok\}$
Work Step by Step
$sin(10\theta)=\frac{\sqrt{3}}{2}$
$10\theta=sin^{-1}(\frac{\sqrt{3}}{2})$
We know $sin(\theta)$ is positive in quardent $I$ and quardent $II$
The period of the sine function is $360^o$
$10\theta=60^o\;\;\;\;\;\;\;or\;\;\;\;\;\;\;10\theta=180^o-60^o$
$10\theta=60^o+360^ok\;\;\;\;\;\;\;or\;\;\;\;\;\;\;10\theta=120^o+360^ok $
$\theta=6^o+36^ok\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\theta=12^o+36^ok $
$\theta=\{6^o+36^ok\;\;,\;\;12^o+36^ok\},\;\;\;\;\;\;\;\;\;\;when \;\;\;K=\{0,1,2,....\}$
