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}