## Task B-3.2.

Determine all integers $p$ for which the equation $\sin (p x)=\frac{1}{p}$ has 2016 solutions on the interval $\langle 0,2 \pi]$.
## Solution.

The number $p \neq \pm 1$ because $\sin x=1$ has only 1 solution on the given interval. Since $p \in \mathbb{Z}$, $p \neq 0, \pm 1$, the number $\left|\frac{1}{p}\right|<1$, so the equation $\sin x=\frac{1}{p}$ has exactly 2 solutions on the fundamental period of the sine function, i.e., on the interval $\langle 0,2 \pi]$.

Then the equation $\sin (p x)=\frac{1}{p}$ has exactly 2 solutions on its fundamental period, which is $\frac{2 \pi}{|p|}$.

1 point

Since $2 \pi=|p| \cdot \frac{2 \pi}{|p|}$, it follows that the equation $\sin (p x)=\frac{1}{p}$ has exactly $2|p|$ solutions on the interval $\langle 0,2 \pi]$.

1 point

Then from $2|p|=2016$ it follows that $p \in\{-1008,1008\}$.