Answer
$$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$
Work Step by Step
Given $$ \cos 2x \cos x - \sin 2x \sin x=\frac{ \sqrt{2}}{2},\ \ 0\leq x<2\pi $$
Then using $\cos(A+B)= \cos A \cos B - \sin A \sin B$ , we get
\begin{align*}
\cos 2x \cos x - \sin 2x \sin x&=\frac{ \sqrt{2}}{2}\\
\cos(3x)&=\frac{ \sqrt{2}}{2}\\
\end{align*}
First we find all possible solutions for $x$
$$ 3x=\frac{\pi}{4}+2k\pi \ \Rightarrow x= \frac{\pi}{12}+ \frac{2k\pi}{3}$$
To find those solutions that lie in the interval $0\leq x<2\pi$ we let k take on values of 0, 1, and 2; we get
$$x=\frac{\pi}{12},\ \frac{7\pi}{12}, \ \frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{17\pi}{12},\ \frac{23\pi}{12}$$