Answer
$\{\frac{\pi}{12},\frac{\pi}{4},\frac{3\pi}{4},\frac{11\pi}{12},\frac{17\pi}{12},\frac{19\pi}{12}\}$
Work Step by Step
Step 1. Using the Addition Formula for the left side, the equation becomes $sin(2x+x)=\frac{\sqrt 2}{2}$ or $sin(3x)=\frac{\sqrt 2}{2}$ which gives $3x=2k\pi+\frac{\pi}{4}$ and $3x=2k\pi+\frac{3\pi}{4}$ or $x=\frac{2k\pi}{3}+\frac{\pi}{12}$ and $x=\frac{2k\pi}{3}+\frac{\pi}{4}$ where $k$ is an integer.
Step 2. For $x=\frac{2k\pi}{3}+\frac{\pi}{12}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12} $
Step 3. For $x=\frac{2k\pi}{3}+\frac{\pi}{4}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{4},\frac{11\pi}{12},\frac{19\pi}{12} $
Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{12},\frac{\pi}{4},\frac{3\pi}{4},\frac{11\pi}{12},\frac{17\pi}{12},\frac{19\pi}{12}\}$
