5. Solve the equation $\sin \frac{\pi n}{12} \cdot \sin \frac{\pi k}{12} \cdot \sin \frac{\pi m}{12}=\frac{1}{8}$. Here $k, m, n-$ are natural numbers not exceeding 5.
# Solution.

Let $1 \leq n \leq k \leq m \leq 5$.

Consider the cases:

1) $n=k=m=2$. Clearly, this set is a solution.
2) $\quad n=1$. Then note that for $k=2, m=5$ we get a valid equality.

The function $y=\sin x$ is increasing on the interval $\left(0 ; \frac{\pi}{2}\right)$, so the sets $\left.2 ; k ; m\right)$ for $k \geq 3$ and $(1 ; 2 ; 3),(1 ; 2 ; 4) ;(1 ; 3 ; 5),(1 ; 4 ; 5)$ are not solutions.

It remains to verify that $(1 ; 3 ; 4)$ is not a solution.

Answer: $(2 ; 2 ; 2),(1 ; 2 ; 5),(1 ; 5 ; 2),(2 ; 1 ; 5),(2 ; 5 ; 1),(5 ; 1 ; 2),(5 ; 2 ; 1)$.