Convolution of functions $f,g\in L^1([0,1])$ Problem: convolution ($f*g) $ of functions $f,g\in L^1([0,1])$, where:
$$f(x) = \frac{3}{5-4\cos{4\pi x}},$$
$$g(x) = \frac{2\cos{2\pi x}}{5-4\cos{4\pi x}},$$
and $$(f * g)(x) = \int_{0}^{1}f(x-y)g(y)dy.$$
I'm not sure how to approach this problem. This is a problem from a course dealing with Fourier Analysis and I'm working in $[0,1]$ where Fourier series of functions are of the form $$\sum_{n \in\mathbb{Z}} f_c(n) e^{2\pi i nx}$$
where $f_c(n)=\int_{0}^1 f(x)e^{-2\pi i nx}dx.$
What I know is $(f*g)_c (n) = f_c(n)g_c(n),$ but I'm not sure if this would help. We haven't yet dealt with the question of whether a Fourier series converges to the function, unless $f$ is literally equal to the Fourier series itself though, so I'm not sure if that would be helpful at all. 
I also know the result is $0$, and direct integration hasn't yielded any results. 
Any hints would be appreciated! 

First note a couple of translation symmetries:
$$
f(x-1/2) = \frac{3}{5-4\cos{4\pi (x-1/2)}} = \frac{3}{5-4\cos{4\pi x}} = f(x) \\
g(x+1/2) = \frac{2\cos{2\pi (x+1/2)}}{5-4\cos{4\pi (x+1/2)}} = \frac{-2\cos{2\pi x}}{5-4\cos{4\pi x}} = -g(x+1/2)
$$
Then we split the integral into two parts and translate the second one:
$$\begin{align}
(f * g)(x) 
&= \int_{0}^{1} f(x-y) \, g(y) \, dy \\
&= \int_{0}^{1/2} f(x-y) \, g(y) \, dy + \int_{1/2}^{1} f(x-y) \, g(y) \, dy \\
&= \{ z := y - 1/2 \} \\
&= \int_{0}^{1/2} f(x-y) \, g(y) \, dy + \int_{0}^{1/2} f(x-z-1/2) \, g(z+1/2) \, dz \\
&= \int_{0}^{1/2} f(x-y) \, g(y) \, dy + \int_{0}^{1/2} f(x-z) \, (-g(z)) \, dz \\
&= 0.
\end{align}$$
