8.143. $\cos 4 x+2 \cos ^{2} x=1$.

8.143. $\cos 4 x+2 \cos ^{2} x=1$.
## Solution.

We have

$$
\begin{aligned}
& \cos 2(2 x)+2 \cos ^{2} x-1=0, \Leftrightarrow 2 \cos ^{2} 2 x-1+\cos 2 x=0 \\
& 2 \cos ^{2} 2 x+\cos 2 x-1=0
\end{aligned}
$$

Solving this equation as a quadratic in terms of $\cos 2 x$, we find $(\cos 2 x)_{1}=-1, 2 x_{1}=\pi+2 \pi k, x_{1}=\frac{\pi}{2}+\pi k, k \in \mathbb{Z}; (\cos 2 x)_{2}=\frac{1}{2}$, $2 x_{2}= \pm \frac{\pi}{3}+2 \pi n, x_{2}= \pm \frac{\pi}{6}+\pi n, n \in \mathbb{Z}$. Combining $x_{1}$ and $x_{2}$, we get $x=\frac{\pi}{6}+\frac{\pi k}{3}=\frac{\pi}{6}(2 k+1), k \in \mathbb{Z}$.

Answer: $\quad x=\frac{\pi}{6}(2 k+1) \quad k \in \mathbb{Z}$.