$\theta = \dfrac{n\pi}{2}$ 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.

  • $\begingroup$ Better to use words in your title instead of using Latex only. $\endgroup$ – StubbornAtom Apr 10 '19 at 15:00
  • $\begingroup$ @StubbornAtom can you please suggest a better title? Or give an example. $\endgroup$ – user541396 Apr 10 '19 at 15:02
  • $\begingroup$ As you can see I edited your title with some formatting adjustments. You can see this in the edit history. Related threads on Meta: math.meta.stackexchange.com/questions/9687/…, math.meta.stackexchange.com/questions/8891/…. $\endgroup$ – StubbornAtom Apr 10 '19 at 15:05
  • $\begingroup$ @StubbornAtom titles are supposed to be short so I usually don't add details like range of $\theta$ and unnecessary language. But I will take care of your suggestion next time. $\endgroup$ – user541396 Apr 10 '19 at 15:11



$\dfrac{4(m-1)}5=n-1$ 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?

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  • $\begingroup$ Can you please elaborate on the last step? $\endgroup$ – user541396 Apr 10 '19 at 14:36
  • $\begingroup$ Do you mean why $m-1=5r?$ $\endgroup$ – lab bhattacharjee Apr 10 '19 at 14:38
  • $\begingroup$ 4(m-1) should be a multiple of five , so. 4(m-1) = 5r where r is a integer..... I got that. But what happened next ? $\endgroup$ – user541396 Apr 10 '19 at 14:45
  • $\begingroup$ But $4,5$ don't have a common factor$>1$ $\endgroup$ – lab bhattacharjee Apr 10 '19 at 14:46
  • $\begingroup$ thank you got that . $\endgroup$ – user541396 Apr 10 '19 at 14:51

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