Answer
$\frac{\pi}{12},\frac{7\pi}{12},\frac{3\pi}{4},\frac{5\pi}{4},\frac{17\pi}{12},\frac{23\pi}{12} $
Work Step by Step
1. Given $\sqrt 2cos(3x)-1$, we have $cos(3x)=\frac{\sqrt 2}{2}$
2. For $x\in[0,2\pi)$ and $3x\in[0,6\pi)$, we can find $3x=\frac{\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4},\frac{15\pi}{4},\frac{17\pi}{4},\frac{23\pi}{4} $
3. Thus $x=\frac{\pi}{12},\frac{7\pi}{12},\frac{3\pi}{4},\frac{5\pi}{4},\frac{17\pi}{12},\frac{23\pi}{12} $
