Find all $x$ such that lim $\sin (nx)$ exists

Question

I know how to prove that the set of $x$ such that $\lim_{n\to\infty} \sin(nx)$ exists has measure zero. And clearly the limit exists when $x = \pi k$ for $k\in\mathbb{Z}$. I'm guessing those are the only such $x$, but how do I prove that?

Answer 1

if $x\neq Q\pi$, then the set of the rest of division of $nx$ by $2\pi$ is dense in $[0,2\pi]$ so no limit.

If $x=p/q\pi$, and $x\neq k\pi$, $\sin(nqx)=0$ but $\sin(2nqx +x)=sin(x)\neq 0$ so no limit again

Answer 2

$\sin(nx)=\text{Im}\left(e^{inx}\right)$. If $\frac{x}{\pi}$ is an irrational number, $e^{inx}$ is dense in the unit circle, hence $\sin(nx)$ is dense in the interval $[-1,1]$ (projections preserve density). So, in order that $\sin(nx)$ has a limit, $\sin(nx)$ has to be a periodic sequence. The only periodic and convergent sequences are the constant ones: $x\in\pi\mathbb{Z}$ readily follows.