I need to show that $\{\cos(nx)\}_{n=0}^{\infty}$ is a complete system in $L_2[0, \pi]$.
My idea was to declare $g(x):=\begin{cases} f(x) &\text{if } x\in [0,\pi] \\{}\\ f(-x) &\text{ if } x\in [-\pi,0]\end{cases}$.
Since $g \in L_2[-\pi, \pi]$ and $\{1, \cos(nx), \sin(nx)\}_{n=1}^{\infty}$ is a complete system,
then for every $\epsilon > 0 $ there is $ h \in\operatorname{span}(\{1, \cos(nx), \sin(nx)\}_{n=1}^{\infty})$ such that $\|g(x) - h(x)\|_2 < \epsilon$.
Now, $h(x) = \sum_{n=0}^{N}a_n\cos(nx) + \sum_{n=1}^{N}b_n\sin(nx)$ for some $N \in \mathbb{N}$.
From the above I get $g(x) = \sum_{n=0}^{\infty}a_n\cos(nx) + \sum_{n=1}^{\infty}b_n\sin(nx)$ , and since $g(x) = g(-x)$ it follows that $0 = 2\sum_{n=1}^{\infty}b_n\sin(nx)$.
Here is where I got stuck. I want to say $b_n = 0$ but $\sum_{n=1}^{\infty}b_n\sin(nx) = 0$ doesn't guarantee that (for example $b_n = \frac{1}{n}$).
Help would be very appreciated!