2. Determine the sum of all solutions of the equation $\log _{8}(\sin 2 x)=-\frac{1}{3}$ in the interval $[0,2 \pi]$.

(8)
Solution. Given the condition $\sin 2 x>0$, from $\log _{8}(\sin 2 x)=-\frac{1}{3}=\log _{8} 8^{-\frac{1}{3}}=\log _{8} \frac{1}{2}$ we obtain

$$
\sin 2 x=\frac{1}{2}
$$

which satisfies the condition and gives us

$$
2 x=\frac{\pi}{6}+2 k \pi \quad \text { or } \quad 2 x=\frac{5 \pi}{6}+2 k \pi
$$

i.e.,

$$
x=\frac{\pi}{12}+k \pi \quad \text { or } \quad x=\frac{5 \pi}{12}+k \pi
$$

On the interval $[0,2 \pi]$, the solutions are $x_{1}=\frac{\pi}{12}, x_{2}=\frac{5 \pi}{12}, x_{3}=\frac{13 \pi}{12}, x_{1}=\frac{17 \pi}{12}$.

The sum of all solutions is $\frac{\pi}{12}+\frac{5 \pi}{12}+\frac{13 \pi}{12}+\frac{17 \pi}{12}=3 \pi$.

(full solution 8 points)

(partial solution $(*) 4$ points)