Answer
Stationary points $(1+2 k) \frac{\pi}{2}=x$ where $k \in \mathbb{Z}$
Critical Points $(1+2 k) \frac{\pi}{2}=x$ where $k \in \mathbb{Z}$ and $x=0$
Work Step by Step
When $x<0 ; f(x)=\sin (-x)=-\sin x$
When $x>0 ; f(x)=\sin x$
Let $f_{1}(x)=\sin x ; f_{2}(x)=-\sin x$
$f_{1}^{\prime}(x)=\cos x ; f_{2}^{\prime}(x)=-\cos x$
Thus in $x>0,and \ \cos x=0$ at $(-1+2 k) \frac{\pi}{2}$ where $k \in \mathbb{Z}^{+}$
In $x<0; finally, \ -\cos x=0$ at $(1+2 k) \frac{\pi}{2}$ where $k \in \mathbb{Z}$
When we combine these results and generalize for all the values of $x$ , we have that the and stationary points of the function are $x=(1+2 k) \frac{\pi}{2}$ where $k \in \mathbb{Z}$
The critical points also include the value of $x$ where the function is not differentiable.
$x=0$
