Suppose $f_1\in L^{3/2}([0,1]),f_2\in L^3([0,1])$. Then $\lim_{n\to \infty} \int_0 ^1 f_1(x)f_2(x)\cos (2\pi nx)=0$
Idea 1: Since I see $L^p$ functions in an integral, Holder's inequality comes to mind. $3/2,3$ are Holder conjugates since $1/(3/2)+1/3=1$. Applying Holder, $\vert \int_X f_1(x)f_2(x)\cos (2\pi nx)\vert\le \int_X \vert f_1(x)f_2(x)\cos (2\pi nx)\vert \le \Vert f_1\Vert _{3/2} \Vert f_2\Vert_{3}$, but I can't see why this upper bound would tend to 0. It was probably too naive to just replace cosine with its upper bound of 1.
Idea 2: Since I have a periodic function in my integrand, the Riemann-Lebesgue Lemma comes to mind but I only know how to use this when the non-periodic part of the integrand is in $L^1$.
How can I show the limit of these integrals is $0$?
Your second idea works fine: just observe that the function $x\to f_1(x)f_2(x)$ is $L^1$ (you proved that with Hölder's indequality).