B2. Solve the equation $\log _{2}\left(2 \cos ^{2} x-2 \cos x+1\right)=-1$.
B2. Let's write $y=2 \cos ^{2} x-2 \cos x+1$. Then $\log _{2} y=-1$, so $y=2^{-1}=\frac{1}{2}$. It follows that $\frac{1}{2}=2 \cos ^{2} x-2 \cos x+1$ or $(2 \cos x-1)^{2}=0$. Therefore, $\cos x=\frac{1}{2}$ and thus $x= \pm \frac{\pi}{3}+2 k \pi$ for $k \in \mathbb{Z}$.