8.291. $\cos 10 x+2 \cos ^{2} 4 x+6 \cos 3 x \cos x=\cos x+8 \cos x \cos ^{3} 3 x$.
## Solution.

We have:

$\cos 10 x+(1+\cos 8 x)+3(2 \cos 3 x \cos x)=\cos x+2(2 \cos x \cos 3 x) \cos ^{2} 3 x \Leftrightarrow$ $\Leftrightarrow \cos 10 x+1+\cos 8 x+3(\cos 2 x+\cos 4 x)=$

$=\cos x+2(\cos 2 x+\cos 4 x) \cdot(1+\cos 6 x)$

$\cos 10 x+1+\cos 8 x+3 \cos 2 x+3 \cos 4 x=\cos x+2 \cos 2 x+$

$+2 \cos 4 x+2 \cos 2 x \cos 6 x+2 \cos 4 x \cos 6 x \Leftrightarrow$

$\Leftrightarrow \cos 10 x+1+\cos 8 x+\cos 2 x+\cos 4 x=$

$=\cos x+\cos 4 x+\cos 8 x+\cos 2 x+\cos 10 x \Rightarrow$

$\Rightarrow \cos x=1, x=2 \pi k, k \in Z$.

Answer: $x=2 \pi k, k \in Z$.