I think your solution is incomplete, because if we have $x_1-x_2=2\pi$, and $f(x_1)=0$, the given equality must hold. An easy way how to solve it is to put $f(x)=r.cos(x+a)$, it's easz to see we can do it. Than we have $r.cos(x_1+a)=0$, which means $x_1+a=\frac{\pi} 2 +k\pi$, similarly $x_2+a=\frac{\pi} 2 +l\pi$ from where $x_1-x_2=(k-l)\pi$.

Kondr wrote: I think your solution is incomplete, because if we have $x_1-x_2=2\pi$, and $f(x_1)=0$, the given equality must hold. An easy way how to solve it is to put $f(x)=r.cos(x+a)$, it's easz to see we can do it. Than we have $r.cos(x_1+a)=0$, which means $x_1+a=\frac{\pi} 2 +k\pi$, similarly $x_2+a=\frac{\pi} 2 +l\pi$ from where $x_1-x_2=(k-l)\pi$. I dont know why we can set $f(x)=r\cdot\cos (a+x)$ , it is not easy for me to see it though . Another thing , isnt the general solution for $\cos (a+x_1)=0$ is $x_1+a=2\pi k\pm \frac{\pi}{2}$ where $k$ is integer .

To the first thing: Let there be n vectors $v_1,v_2,...,v_n$, such that thez begin in [0,0], they have lenghts $1$, $\frac 1 2$, $\frac 1 4$ etc. and angles with x-axis $a_1, a_2, a_3, ....$. Then the given sum of expression with cos is a sum of x-coordinates of these vectors rotated by angle x. But this is (i hope obviously) equal to the x-coordinate of vector $V=v_1+v_2+...+v_n$ rotated by angle x. If we denote $r$ the lenght of $V$ and $a$ the angle of V and x-axis, the result follows. To be precise, we would have to show that $r>0$, but from triangle inequality we have $r\geq 1-\frac 1 2 -\frac 1 4-...-\frac 1 {2^n}=\frac 1 {2^n}$ To the second thing: As far as I know $cos\left(\frac 3 2 \pi \right)=0$.

$f(x)$can be written as $ r.cos(x_{1}+a)=0 $ because differentiating $f(x)$ twice, you get $-f(x)$, which happens only when it's a $sin$ or $cos$. And the powers of $2$ in the coefficient are arbitrary in this problem.

golopot wrote: $f(x)$can be written as $ r.cos(x_{1}+a)=0 $ because differentiating $f(x)$ twice, you get $-f(x)$, which happens only when it's a $sin$ or $cos$. And the powers of $2$ in the coefficient are arbitrary in this problem. But you should also show that $r\ne 0$ in order to get the required final result. And I think that the powers of two in the coefficient are useful for that forgotten part.

I think phasors can be used for solving this. similar to Kondr's idea