Answer
$\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8} \}$
Work Step by Step
Step 1. Using the identity $sin(2x)=2sin(x)cos(x)$, we have $sin(2x)=\frac{\sqrt 2}{2}$, which gives $2x=2k\pi+\frac{\pi}{4}$ and $2x=2k\pi+\frac{3\pi}{4}$ or $x=k\pi+\frac{\pi}{8}$ and $x=k\pi+\frac{3\pi}{8}$ where $k$ is an integer.
Step 2. For $x=k\pi+\frac{\pi}{8}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{8},\frac{9\pi}{8}$
Step 3. For $x=k\pi+\frac{3\pi}{8}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{3\pi}{8},\frac{11\pi}{8}$
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8} \}$
