## Precalculus (6th Edition) Blitzer

$2k\pi+\frac{\pi}{4}$, $2k\pi+\frac{3\pi}{4}$
Step 1. Given the equation $sin(x)=\frac{\sqrt 2}{2}$, we have to find the solutions in $[0,2\pi)$ as $x_1=\frac{\pi}{4}$ and $x_2=\frac{3\pi}{4}$ Step 2. Consider that the function has a period of $2\pi$; we can express the solutions as $x=2k\pi+\frac{\pi}{4}$ and $x=2k\pi+\frac{3\pi}{4}$ where $k$ is an integer.