Answer
Stationary points $x= \frac{\pi}{2}(1+2 k)$ where $k \in \mathbb{Z}$
Critical Points $\mathrm{x}=k \pi$ and $x=\frac{\pi}{2}(1+2 k) $ where $k \in \mathbb{Z}$
Work Step by Step
Given $|\sin x|=f(x)$
When $x \in(\pi, 2 \pi) ; -\sin x=f(x)$
When $x \in(0, \pi) ;\sin x= f(x)$
Let $\sin x=f_{1}(x)$
$\cos x=f_{1}^{\prime}(x)$
$f_{2}(x)=-\sin x$
$f_{2}^{\prime}(x)=-\cos x$
Thus in $(0, \pi), \cos x=0$ at $\frac{\pi}{2}$
$\operatorname{In} (\pi, 2 \pi),-\cos x=0$ at $\frac{3 \pi}{2}$
When we combine these results and generalize for all the values of x, we have that the and stationary points of the function are $(1+2 k) \frac{\pi}{2}=x$ where $k \in \mathbb{Z}$
The critical points also include the value of $x$ where the function is not differentiable.
And then $k \pi=\mathrm{x}$ where $k \in \mathbb{Z}$
