The function $f'+f'''$ has at least $3$ zeros on $[0,2\pi]$. 
Show that if $f\in \mathcal{C}^3$ and $2\cdot\pi$ periodic then the function $f'+f'''$ has at least $3$ zeros on $[0,2\pi]$.

My attempt :
f is $2\pi$ periodic and $\mathcal{C}^3$, we have :
$$\lim_{h\rightarrow 0, h>0} \frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h>0} \frac{f(2\pi+h)-f(2\pi)}{h}\Rightarrow f'(0)=f'(2\pi)$$
After that I tried to compute differently the limit 
$$
\lim_{h\rightarrow 0, h>0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h<0}\frac{f(h)-f(0)}{|h|}=\lim_{h\rightarrow 0, h<0}-\frac{f(2\pi+h)-f(0)}{h}\Rightarrow f'(0)=-f'(2\pi) 
$$
wich is clearly false because I get $f'(0)=0$ ( $f(x)=\sin(x)$ is a counterexample).
For $2$ zeros is relatively easy but I am stuck for the additional zero.
EDIT : I find this exercice (as usual) here : Revue de la filière Mathématique. This was asked during an oral examination of École normale supérieure rue d'Ulm. 

A possible direction:
Write out the Fourier series:
$$f(x)= a_0 + \sum\left( a_n \cos nx + b_n \sin nx\right)$$
$$g(x)\equiv f'(x)+f'''(x) = -\sum n (n^2-1) \left(b_n \cos (n x)-a_n \sin (n x)\right)$$
Clearly, when $n=1$, $g(x)=0$, meaning that the lowest order in $g(x)$ is $n=2$. Now, since there is no constant term, by the mean value theorem $g(x)$ must cross zero at at least one point, $x_0$.
Intuitively, any function with no frequencies lower than two, must cross zero at least $4$ times in the interval $[0,2\pi]$, though I'm not sure how to complete this proof.
