8.350. $\cos ^{2}\left(2 x+\frac{\pi}{3}\right)+\cos ^{2}\left(\frac{\pi}{12}-x\right)=0$.
Solution.

Let's write the equation as

$$
\begin{aligned}
& \cos ^{2}\left(2 x+\frac{\pi}{3}\right)+\cos ^{2}\left(\frac{\pi}{12}-x\right)=0 \Leftrightarrow\left\{\begin{array}{l}
\cos \left(2 x+\frac{\pi}{3}\right)=0 \\
\cos \left(x-\frac{\pi}{12}\right)=0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 x + \frac { \pi } { 3 } = \frac { \pi } { 2 } + \pi k , } \\
{ x - \frac { \pi } { 1 2 } = \frac { \pi } { 2 } + \pi l , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{\pi}{12}+\frac{\pi k}{2} \\
x=\frac{7 \pi}{12}+\pi l
\end{array}\right.\right.
\end{aligned}
$$

From this,

$$
\begin{aligned}
& \frac{\pi}{12}+\frac{\pi k}{2}=\frac{7 \pi}{12}+l, k=1+2 l, k \in Z, l \in Z \\
& x=\frac{\pi}{12}+\frac{\pi(1+2 l)}{2}=\frac{7 \pi}{12}+\pi l, l \in Z
\end{aligned}
$$

Answer: $x=\frac{7 \pi}{12}+\pi l, l \in Z$.