5. The solution can only exist if $a \in\left\{-\frac{\pi}{12}\right\} \cup\left(0 ; \frac{\pi}{12}\right]$, since otherwise the left side of the equation is either undefined or strictly positive. When $a=-\frac{\pi}{12}$, the equation becomes $2|x-16|=0$. Therefore, when $a=-\frac{\pi}{12}$, $x=16$. If $a \in\left(0 ; \frac{\pi}{12}\right]$, then $2^{\frac{1}{\sin ^{2}(2 a)}}>16$, and $2^{-4 \operatorname{tg}(3 a)}<1$. Therefore, the minimum value of the function $f(x)=\left|x-2^{\frac{1}{\sin ^{2}(2 a)}}\right|+\left|x-2^{-4 \operatorname{tg}(3 a)}\right|$ is not less than 15. On the other hand, the absolute value of the expression $g(a)=a\left(a+\frac{\pi}{12}\right)^{2}\left(a-\frac{\pi}{12}\right)$ on the half-interval $\left(0 ; \frac{\pi}{12}\right]$ is certainly no more than $1:|g(a)|<a\left(a+\frac{\pi}{12}\right)^{3}<1$.
Therefore, when $a \in\left(0 ; \frac{\pi}{12}\right]$, there are no solutions.
Answer: $x=16$ when $a=-\frac{\pi}{12}$. For other $a$, there are no solutions. Answer to option $5-2: x=-16$ when $a=\frac{\pi}{12}$. For other $a$, there are no solutions.

Lomonosov Moscow State University

## Olympiad "Conquer Sparrow Hills"

Option $6-1$ (Nizhny Novgorod)