We will assume that $0<\pi x<\frac{\pi}{2}$ or $0<x<\frac{1}{2}$
case 1:
If $a$ is a perfect square, say $a=\alpha^2$, $\alpha\in \mathbb{N}^*$, then the question is equivalent to find $x$ rational such that $\cos(\pi x)$ is rational:
$\cos(\pi x)=\frac{\alpha + b}{c}$
Using Niven's theorem, as discussed here or here , the only solution for $0<x<\frac{1}{2}$ is $x=\frac{1}{3} $ ;
$\cos(\frac{\pi}{3})=\frac{1}{2}$
Therefore, $c$ must be even, $a=(\frac{c}{2}-b)^2$ and $c>2b$
case 2:
If $a$ is not a perfect square, here is a conjecture :
there is a unique solution $x=\frac{1}{5}$ and $\cos(\frac{\pi}{5})=\frac{\sqrt{5}+1}{4}$
more precisely: $x=\frac{1}{5}$ ; $ a=5k^2$ ; $b=k$ ; $c=4k$ ; $k\in\mathbb{N}^*$
Any proof or counterexample?
UPDATE
The link given by @Fabio Lucchini proves that the topic is not new and has been already studied in details.
The conjecture is true but more importantly the results are for
$cos(\pi x)=\frac{\sqrt a +b}{c}$ ; $0\leq x\leq \frac{1}{2}$ ; $x\in\mathbb{Q}$ , $a\in \mathbb{N}$ , $b\in \mathbb{N}$ , $c\in \mathbb{N^*}$ i.e including $a=0$ or $b=0$
$x=0 $ ; $\; $ $\cos(0)=1$
$x=\frac{1}{6}$ ; $\; \cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$
$x=\frac{1}{5}$ ; $\; \cos(\frac{\pi}{5})=\frac{\sqrt 5 +1}{4}$
$x=\frac{1}{4}$ ; $\; \cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}$
$x=\frac{1}{3}$ ; $\; \cos(\frac{\pi}{3})=\frac{1}{2}$
$x=\frac{1}{2}$ ; $\; \cos(\frac{\pi}{2})=0$
