Given that $f: \mathbb{R} \to \mathbb{R}$ is differentiable and $f'(x)=f(x+\frac{\pi}{2})$ for all $x\in \mathbb{R}$, prove that there exists $x_0$ such that $f(x_0)=0$
I thought that $f(x)=\cos(x)$ but I don't know how to prove it. Moreover is there a solution without using the $\cos(x)$?
Let us consider the equation $$ f'(x) = f(x+a) $$ with $a \ge 1$ instead. Assume that $f$ has no zeros. Without loss of generality we can assume that $f$ is strictly positive.
It follows that $f$ and $f'$ are strictly increasing, so that for $x < y$ $$ \tag {*} f(y) > f(x) + (y-x) f'(x) \, . $$ In particular, $\lim_{x \to \infty} f(x) = +\infty$.
On the other hand, using $(*)$ again, $$ f'(x) = f(x + a) > f(x) + a f'(x) $$ implies $$ (a-1)f'(x) + f(x) < 0 $$
If $a=1$ then $f(x) < 0$ gives an immediate contradiction. If $a > 1$ then the last inequality shows that $$ e^{x/(a-1)} f(x) $$ is decreasing, so that $$ f(x) \le f(0) e^{-x/(a-1)} \, , $$ contradicting the fact that $\lim_{x \to \infty} f(x) = +\infty$.
Remark: If $0 < a \le 1/e$ then $e^{\lambda a} = \lambda$ has a real solution $\lambda$, and $f(x) = e^{\lambda x}$ satisfies the functional equation and has no zeros (compare How to solve differential equations of the form $f'(x) = f(x + a)$).