I'm trying to find what is the set of all rational numbers $r\in\mathbb{Q}$ s.t. $\cos(r\cdot\pi)\in\mathbb{Q}$. I have instructions saying I should look at $z=\cos q\pi+i\sin q\pi\in\mathbb{C}$, and note that it's a root of $x^{2n}-1$, and then look at the polynomial $$f\left(x\right)=\left(x-z_{q}\right)\left(x-\overline{z_{q}}\right)=x^{2}-2\cos\left(q\pi\right)x+1$$
I assume the solution uses tools such as Gauss's Lemma.
I've seen answers on SE (such as this one), and they all give valid answers but they use tools such as "algebraic integers" (and another solution used a simple method without any advanced math), but as this is a question from (a past exam in) an "Abstract Algebra" course we don't learn about algebraic numbers, I'd like to know what is the way to solve it using these instructions (without theorems regarding algebraic integer, and probably using theorems about reducible polynomials such as Gauss's Lemma).
You know that roots $f(x)$ are also roots of $x^{2n}-1$, where $n$ is the denominator of $q$. Therefore $f(x)$ divides $x^{2n}-1$, so you can write $f(x) q(x) = x^{2n}-1$ for some $q \in \mathbb{Q}[x]$. Now you can finish the solution using Gauss's lemma.