4. First, we solve the equation:

$$
\begin{aligned}
& \cos 4 x-6 \cos ^{2} 2 x+8 \cos ^{2} x=0 \Longleftrightarrow \\
& 2 \cos ^{2} 2 x-1-6 \cos ^{2} 2 x+4(1+\cos 2 x)=0 \Longleftrightarrow \\
& 4 \cos ^{2} 2 x-4 \cos 2 x-3=0 \Longleftrightarrow \\
& \cos 2 x=-1 / 2 \Longleftrightarrow \\
& x= \pm \frac{\pi}{3}+\pi n, \quad n \in \mathbb{Z}
\end{aligned}
$$

Constraints related to the denominator: $x \in(1 ; 5)$. Since $1 < \frac{\pi}{3} < 2 < \frac{2 \pi}{3} < 3 < \frac{4 \pi}{3} < 4 < \frac{5 \pi}{3} < 5$, then $x=\pi / 3 ; 2 \pi / 3 ; 4 \pi / 3$.
Answer: $\pi / 3 ; 2 \pi / 3 ; 4 \pi / 3$. Answer to option: $7-2: \pi / 3 ; 2 \pi / 3 ; 4 \pi / 3$.