Good insights to start off about the periodicity of the sine function.
Let's think about this in parts - dissecting the function from the outside inward.
Remember the definition of absolute value:
$$f(x) = | g(x) | = \left\{ \begin{array}{ll}
g(x) & g(x) \geq 0 \\
-g(x) & g(x) < 0
\end{array} \right.$$
$$ g(x)=\sin(|x|)=\sin\left( \begin{array}{ll}
x & x \geq 0 \\
-x & x < 0
\end{array}\right)$$
By definition, the minima cannot be lower than 0. That does not make it the minima of the function, but it is a good check for the result we get.
So what is the maxima/minima of $\sin(x)$?
$$-1 \leq \sin(x)\leq 1$$
So what is the minima applying the absolute value?
$$ 0 \leq |\sin(x)| \leq 1 \forall x$$
Those are the maxima/minima of the function, regardless of the $x$ input. Let's solve for $x$:
$$\sin x = 0 \rightarrow x = ... ,-2\pi,\pi,0,\pi,2\pi, ...$$
$$\sin x = 1 \rightarrow x = ... ,{-3\pi \over 2},{-\pi \over 2},{\pi \over 2},{3\pi \over 2},...$$
Notice all the negative values, those don't apply since we apply the absolute value to $x$.
$$x_{min}=0,\pi,2\pi,...$$
$$x_{max}={\pi \over 2},{3\pi \over 2},...$$
There are no local maxima/minima which are not the global maxima/minima.