4. $\{0\}$.

The original inequality can be transformed into
$$(x-\sin x)^{2} \leqslant 0 \Rightarrow x-\sin x=0 \text {. }$$

Construct the function $f(x)=x-\sin x$.
Since $f^{\prime}(x)=1-\cos x \geqslant 0$, the function $f(x)=x-\sin x$ is monotonically increasing on $(-\infty,+\infty)$, and it is easy to see that $f(0)=0$. Therefore, $x-\sin x=0$ has only one real root, 0. Consequently, the solution set of the original inequality is $\{0\}$.