8.182. $8 \cos ^{4} x-8 \cos ^{2} x-\cos x+1=0$.

8.182. $8 \cos ^{4} x-8 \cos ^{2} x-\cos x+1=0$. 

(No change needed as the text is already in English and is a mathematical equation which is universal.)
## Solution.

Rewrite the equation as $8\left(\cos ^{2} x\right)^{2}-8 \cos ^{2} x-\cos x+1=0$.

$$
\begin{aligned}
& 8\left(\frac{1}{2}(1+\cos 2 x)\right)^{2}-4(1+\cos 2 x)-\cos x+1=0 \Leftrightarrow 2(1+\cos 2 x)^{2}- \\
& -4(1+\cos 2 x)-\cos x+1=0 \Leftrightarrow \\
& \Leftrightarrow 2+4 \cos 2 x+2 \cos ^{2} 2 x-4-4 \cos 2 x-\cos x+1=0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos ^{2} x-1-\cos x=0 \Leftrightarrow \cos 4 x-\cos x=0 \Leftrightarrow \\
& \Leftrightarrow 2 \sin \frac{4 x+x}{2} \sin \frac{x-4 x}{2}=0,-\sin \frac{5 x}{2} \sin \frac{3 x}{2}=0
\end{aligned}
$$

From this, either

$$
\sin \frac{5 x}{2}=0, \frac{5 x}{2}=\pi n, x_{1}=\frac{2}{5} \pi n, n \in Z
$$

or

$$
\sin \frac{3 x}{2}=0, \frac{3 x}{2}=\pi k, x_{2}=\frac{2}{3} \pi k, k \in Z
$$

Answer: $x_{1}=\frac{2}{5} \pi n ; x_{2}=\frac{2}{3} \pi k$, where $n$ and $k \in Z$.