1. Determine all real numbers $x$ for which

$$
\sin x=\sin 2x=\sin 3x=\cdots=\sin 2017x
$$
1. From the first equality, we have $\sin x=2 \sin x \cos x$, i.e., $\sin x(1-2 \cos x)=0$, from which we conclude $\sin x=0$ or $\cos x=\frac{1}{2}$. In the first case, we have $x \in\{k \pi : k \in \mathbb{Z}\}$, and we see that all these values indeed satisfy all the equalities in the problem statement. In the second case, we have $x= \pm \frac{\pi}{3}+2 k \pi$, but then $\sin x \neq 0$ while $\sin 3 x=\sin ( \pm \pi+6 k \pi)=0$, so the sequence of equalities in the problem statement is not satisfied. Therefore, the solution to the equation is $x \in\{k \pi \mid k \in \mathbb{Z}\}$.