Answer
$x=\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\}$
Work Step by Step
$sin(2x)=\frac{\sqrt{2}}{2}$
$2x=sin^{-1}(\frac{\sqrt{2}}{2})$
We know $sin(x)$ is positive in quardent $I$ and quardent $II$
The period of the sine function is $2\pi$
$2x=\frac{\pi}{4}\;\;\;\;or\;\;\;\;\;\;2x=2\pi+\frac{\pi}{4}\;\;\;\;or\;\;\;\;\;\;2x=\pi -\frac{\pi}{4}\;\;\;\;\;\;or\;\;\;\;2x=2\pi + \frac{3\pi}{4} $
$x=\frac{\pi}{8}\;\;\;\;or\;\;\;\;\;\;x=\frac{9\pi}{8}\;\;\;\;or\;\;\;\;\;\;x=\frac{3\pi}{8}\;\;\;\;\;\;\;\;or\;\;\;\;\;x=\frac{11\pi}{8}$
$x=\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\}$
