How to make sure that solutions of an equation are rational multiple of $\pi$ or not?

Question

I have this two-variable equation $$\frac{\left(3 y^2+1\right)^2 \left(2 \cosh \frac{26 \pi y}{15} \cosh x y-\cosh (2 \pi -x) y\right)}{9 \left(y^2-1\right)^2 \cosh (x+2 \pi ) y+8 \left(3 y^2-1\right) \cosh x y}=1$$ with $0<x,y<3$. Is there a way to make sure whether those $x$ and $y$ which solve this equation are rational multiple of $\pi$ or not?

Answer 1

The definition of a number rational is represented by the division of two integers, so we have:

$ \frac{a}{b} $

where a and b are integers, so if to get any number rationals multiples of pi, we will have:

$ \frac{a}{b} \cdot \pi $

So we want to make sure that x and y will be rational multiple of pi, so we just need to substitute x and y to a rational multiple of pi, so if we assume $ x = \frac{a}{b} \cdot \pi $ and $ y = \frac{d}{e} \cdot \pi $ we will have this:

$ \frac{(3(\frac{a}{b}\cdot \pi)^2 + 1)^2(2 \cdot \cosh{\frac{a}{b}\cdot \pi \frac{d}{e} \cdot \pi} - \cosh(2\pi - \frac{a}{b}\cdot \pi)\frac{d}{e} \cdot \pi)}{9((\frac{d}{e} \cdot \pi)^2 - 1)^2 \cosh(\frac{a}{b}\cdot \pi + 2\pi)y + 8(3(\frac{d}{e} \cdot \pi)^2 - 1)\cosh{\frac{a}{b}\cdot \pi\frac{d}{e} \cdot \pi}} = 1 $

with $ 0 < \frac{a}{b} \cdot \pi,\frac{d}{e} \cdot \pi < 3 $ and $ a,b,c,d,e$ are integers.