Consider rational numbers $\frac{m}{n}$ and $\frac{m'}{n'}$, where $0<\frac{m}{n}, \frac{m'}{n'} <1$.
Then $$\sin^2 (\tfrac{m}{n} \pi) = 2 \sin^2 (\tfrac{m'}{n'} \pi)$$
When $\frac{m}{n} = \frac{1}{4}, \frac{3}{4}$ and $\frac{m'}{n'} = \frac{1}{6}, \frac{5}{6}$. (Both sides equal $\frac{1}{2}$.)
Is it possible that other $\frac{m}{n}$ and $\frac{m'}{n'}$ satisfy these conditions? Is it possible to prove that there are or are not other solutions?