8.328. $\cos ^{2}\left(x+40^{\circ}\right)+\cos ^{2}\left(x-40^{\circ}\right)-\sin 10^{\circ} \cos 2 x=\sin 2 x$.
## Solution.

Reducing the degree, we have

$$
\begin{aligned}
& \frac{1}{2}\left(1+\cos \left(2 x+80^{\circ}\right)\right)+\frac{1}{2}\left(1+\cos \left(2 x-80^{\circ}\right)\right)-\sin 10^{\circ} \cos 2 x=\sin 2 x \\
& 1+\frac{1}{2}\left(\cos \left(2 x+80^{\circ}\right)+\cos \left(2 x-80^{\circ}\right)\right)-\sin 10^{\circ} \cos 2 x=\sin 2 x \Leftrightarrow \\
& \Leftrightarrow 1+\cos 2 x \cos 80^{\circ}-\sin 10^{\circ} \cos 2 x=\sin 2 x \Leftrightarrow \\
& \Leftrightarrow 1+\cos 2 x \sin 10^{\circ}-\sin 10^{\circ} \cos 2 x=\sin 2 x \\
& \sin 2 x=1, 2 x=\frac{\pi}{2}+2 \pi k, x=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1) k \in \mathbb{Z} \\
& \text{Answer: } x=\frac{\pi}{4}(4 k+1), k \in \mathbb{Z}
\end{aligned}
$$