Check $x=0$: Write $h(x)=\sin|x|-|x|$. For $x>0$: $h(x)=\sin x-x$, $h'(x)=\cos x-1$. For $x<0$: $h(x)=-\sin x+x$, $h'(x)=-\cos x+1$. At $x=0$: left limit $=0$, right limit $=0$ — differentiable.

The term $2(x-\pi)\cos|x|$ is clearly differentiable everywhere (product of smooth functions, since $\cos|x|$ is even and smooth).

At $x=\pi$: $\cos|\pi|=\cos\pi=-1$, and all parts are smooth for $x>0$.

Therefore $f$ is differentiable everywhere and $K=\boldsymbol{\emptyset}$.