Answer
$x = \{\frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$
Work Step by Step
$\sqrt{2}~sin~3x-1 = 0$
$sin~3x = \frac{1}{\sqrt{2}}$
$sin~3x = \frac{\sqrt{2}}{2}$
$3x = arcsin(\frac{\sqrt{2}}{2})$
$3x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{17\pi}{4}, \frac{19\pi}{4}$
$x = \frac{\pi}{12}, \frac{3\pi}{12}, \frac{9\pi}{12}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$
$x = \frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}$
For the domain $[0, 2\pi]$, the solution set is as follows:
$x = \{\frac{\pi}{12}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}\}$
