Let $f: \mathbb{R} \longrightarrow \mathbb{R} $ be a differentiable and even function. If $f$ is periodic and the (minimal) period $L>0$, then $f'$ has $2$ zeros in $[0,L)$?
For example, this occurs if we consider $f(x)=\cos(x)$, for all $x \in \mathbb{R} $, since in this case $L=2\pi$.
This is in general true?
No. Define the function $f\colon[-2,2]\to\mathbb{R}$ by \begin{align} f(x)=\begin{cases} -x^2+2\,, &\text{if $|x|\leq 1$}\,, \\ (|x|-2)^2\,, &\text{if $|x|>1$}\,. \end{cases} \end{align} Then $f'(-2)=f'(2)=0$. We can shift $f$ by integral multiples of $T>4$. The resultant graph is a function $F$ as desired with period $T$, such that $F'(x)=0$ for all $x\in[4,T)$.
Edit: Forgot you were asking for the zeroes of $F'$ half way writing. $F'$ must have at least two zeroes, for there must be at least one local maximum and one local minimum in a period.