If $\sin(n!\, x)\to 0$ as $n\to +\infty$, is then $x$ inevitably a rational multiple of $\pi$? If $x$ is a rational multiple of $\pi$, for a natural number $N$ big enough $\sin(n!\,x) = 0$ for all $n\geqslant N$ and then $\sin(n!\,x)\to 0$ as $n\to +\infty$.
However, I'm not so sure about the converse anymore: If it holds that $\sin(n!\,x)\to 0$ as $n\to +\infty$, does it necessarily follow that $x$ belongs to $\pi\mathbf{Q}$?
 A: Here we extend the idea discussed in the comments and discuss the condition for $x$ that makes the limit zero.
Let $x = \pi r$. Write the fractional part $\{r\} = r -\lfloor r \rfloor$ in the form
$$
\{r\} = \sum_{j=2}^{\infty} \frac{a_j}{j!} \quad \text{for} \quad a_j \in \{\,0, 1,\ldots, j-1\,\}.
$$
Then we note that
$$
\begin{split}
|\sin(n!\, \pi r)|
= \Biggl| \sin\biggl(n!\, \pi\sum_{j=2}^{\infty} \frac{a_j}{j!} \biggr) \Biggr|
= \Biggl| \sin\biggl(\pi \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots(n+k)} \biggr) \Biggr|
\end{split}
$$
and
$$
\begin{split}
\frac{a_{n+1}}{n+1}
&\leq \sum_{k=1}^{\infty} \frac{a_{n+k}}{(n+1)(n+2)\cdots (n+k)} \\
&\leq \frac{a_{n+1}}{n+1} +\sum_{k=2}^{\infty} \frac{n+k-1}{(n+1)(n+2)\cdots (n+k)} \\
&= \frac{a_{n+1} + 1}{n+1}.
\end{split}
$$
From this, it is not hard to check that $\sin(n!\, \pi r) \to 0$ if and only if the sequence $(a_{n+1}/(n+1))$ has only limit points in $\{\,0, 1\,\}$, or equivalently, $\operatorname{dist}(a_{n+1}/(n+1), \mathbb{Z})\to 0$. Thomas's example corresponds to the case where all $a_j$ are either $0$ or $1$.
Finally, since the set of all such $r$ is uncountable, there is an irrational number $r$ such that $\sin(n!\, \pi r) \to 0$.
