Just try

One way is to expand $x^2$ in a fourier series which is obviously where the problem is given from, Here is another method a bit based on assumption,

Let us consider $c=2$ , then $\displaystyle f(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n}\cos(nx)}{n^2}\implies f''(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\cos(nx)$

Consider $\cos(nx)$ as the real part of $e^{inx}$ and then a simple G.P sum whose real part becomes $\frac{1}{2}$ .

Now that , $\displaystyle f''(x)=\frac{1}{2}$ with the boundary conditions $f'(0)=0,f(0)=-\bar{\zeta(2)}=-\frac{\pi^2}{12}$ we have by double integrating $\displaystyle f(x)=\frac{x^2}{4}-\frac{\pi^2}{12} \implies x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n\cos(nx)}{n^2}$ as desired.