Answer
$\{0, \frac{\pi}{2}\}$
Work Step by Step
Step 1. Using the identity $sin^2x+cos^2x=1$ and $sin(2x)=2sin(x)cos(x)$ and taking the square on both sides, we have $sin^2x+2sin(x)cos(x)+cos^2x=1$ or $sin(2x)=0$, which gives $2x=2k\pi $ and $2x=2k\pi+\pi$ or $x=k\pi$ and $x=k\pi+\frac{\pi}{2}$ where $k$ is an integer.
Step 2. For $x=k\pi$, we can find all x-values in $[0,2\pi)$ as $x=0,\pi$
Step 3. For $x=k\pi+\frac{\pi}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{2},\frac{3\pi}{2}$
Step 4. Check the answers in original equation; we have solutions in $[0,2\pi)$ as $\{0, \frac{\pi}{2}\}$
