Problem 5. Multiply both sides of the equation by $\sin \frac{x}{2}$ (this transformation is equivalent, as this expression does not turn to zero on the specified segment). Rewrite $2 \sin \frac{x}{2} \sin k x$ as $\cos \frac{2 k-1}{2} x-\cos \frac{2 k+1}{2} x$. After this, all terms will cancel out except the first and the last.

Thus, the roots of the original equation coincide with the roots of the equation

$$
\cos \frac{1}{2} x-\cos \frac{11}{2} x=0
$$

i.e., (converting the difference back to a product) the equation

$$
\sin 3 x \cdot \sin \frac{5}{2} x=0
$$

The roots of the last equation are numbers of the form $\frac{\pi k}{3}$ and $\frac{2 \pi l}{5}$.

It remains to sum the roots that fall within the required interval:

$$
\left(\frac{4 \pi}{3}+\frac{5 \pi}{3}\right)+\left(\frac{6 \pi}{5}+\frac{8 \pi}{5}\right)=\frac{29 \pi}{5}
$$
Answer: $\frac{29 \pi}{5}$.

Remark. $\arccos \frac{-1+\sqrt{5}}{4}=\frac{2 \pi}{5}$ (in particular, the answer $7 \pi-\arccos \left(\frac{-1-\sqrt{5}}{4}\right)-\arccos \left(\frac{-1+\sqrt{5}}{4}\right)$ is also correct).