Answer
$\theta=\{15^o+180^ok\;\;,\;\;75^o+180^ok\}$
Work Step by Step
$sin(2\theta)=\frac{1}{2}$
$2\theta=sin^{-1}(\frac{1}{2})$
We know $sin(\theta)$ is positive in quardent $I$ and quardent $II$
The period of the tangent function is $360^o$
$2\theta=30^o\;\;\;\;or\;\;\;\;\;\;2\theta=180^o-30^o\;\;\;\; $
$2\theta=30^o+360^oK\;\;\;\;or\;\;\;\;\;\;2\theta=150^o+360^oK\;\;\;\; $
$\theta=15^o+180^oK\;\;\;\;or\;\;\;\;\;\;\theta=75^o+180^oK\;\;\;\; $
$\theta=\{15^o+180^ok,75^o+180^ok\},\;\;\;\;\;\;\;\;\;\;when \;\;\;K=\{0,1,2\}$
