Find $\theta\in [0,2\pi]$ if $\theta = \frac{n\pi}{2}$ and $\theta = \frac{2m\pi}{5} + \frac{\pi}{10}$, where $n,m$ are integers

Question

$\theta = \dfrac{n\pi}{2}$ and $\theta = \dfrac{2m\pi}{5} + \dfrac{\pi}{10}$ where $ n ,m \in \mathbb Z$. Find $\theta\in [0,2\pi]$.

It can be solved by hit and try, of course but is there any faster way? Maybe concept of least common multiple can crack it? I tried that but I realised I was only taught to find LCM for natural numbers and not complicated equations. I found a problem which asked me to solve a trigonometric equation and I did until I got stuck here. Thanks a lot.

Answer 1

$\dfrac{n\pi}2=\dfrac{(4m+1)\pi}{10}$

$\iff5n=4m+1\iff5(n-1)=4(m+1)$

$\dfrac{4(m-1)}5=n-1$ which is an integer

$\implies5$ divides $4(m-1)$

$\implies 5$ divides $m-1$ as $(4,5)=1$

WLOG $m-1=5r$ where $r$ is an integer

Can you take it from here?