8.313. $1-\cos x=\sqrt{1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x}}$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}4 \cos ^{2} x-7 \cos ^{4} x \geq 0, \\ 1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x} \geq 0 .\end{array}\right.$

Squaring both sides of the equation, we get

$$
\left\{\begin{array}{l}
1-2 \cos x+\cos ^{2} x=1-\sqrt{4 \cos ^{2} x-7 \cos ^{4} x} \\
1-\cos x \geq 0
\end{array} \Leftrightarrow\right.
$$

$$
\Leftrightarrow\left\{\begin{array}{l}
\sqrt{4 \cos ^{2} x-7 \cos ^{4} x}=2 \cos x-\cos ^{2} x \\
\cos x \leq 1 \\
2 \cos x-\cos ^{2} x \geq 0
\end{array}\right.
$$

Squaring the equation again, we get $4 \cos ^{2} x-7 \cos ^{4} x=4 \cos ^{2} x-4 \cos ^{3} x+\cos ^{4} x$, $8 \cos ^{4} x-4 \cos ^{3} x=0,4 \cos ^{3} x(2 \cos x-1)=0$.

## From this

1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $2 \cos x-1=0, \cos x=\frac{1}{2}, x_{2}= \pm \frac{\pi}{3}+2 \pi n=\frac{\pi}{3}(6 n \pm 1), n \in Z$.

By verification, we confirm that both roots are solutions to the equation.

Answer: $x_{1}=\frac{\pi}{2}(2 k+1) ; x_{2}=\frac{\pi}{3}(6 n \pm 1)$, where $k$ and $n \in Z$.