Answer
$\frac{3\pi}{4}$, $2n\pi+\frac{\pi}{4}$, $2n\pi+\frac{3\pi}{4}$
Work Step by Step
As $sin(x)$ is positive if $x$ is in quadrant I and II, the solutions of $sin(x)=\frac{\sqrt 2}{2}$ in $[0,2\pi)$ are $x=\frac{\pi}{4}$ and $x=\frac{3\pi}{4}$. If $n$ is any integer, all solutions of $sin(x)=\frac{\sqrt 2}{2}$ are given by $x=2n\pi+\frac{\pi}{4}$ and $x=2n\pi+\frac{3\pi}{4}$
