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This question already has an answer here:

Given $x$, $\sin(x) \in \mathbb{Q}$, where $x$ is in degrees, we want to find all $x$ in the range $(0,90)$.

One trivial solution is $x=30$.

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marked as duplicate by Saad, Jyrki Lahtonen Mar 11 '18 at 13:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ That problem asks for both sin as well as cos to be rational. My problem specifically tells that only x (in degrees) and sin(x) need to be rational. $\endgroup$ – kaushal agrawal Aug 21 '17 at 3:28
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If $x$ is a rational multiple of $\pi$ then $e^{ix}$ is a root of unity, and so an algebraic integer. So $2\sin x=-ie^{ix}+ie^{-ix}$ is an algebraic integer. If $\sin x$ is rational too, $2\sin x$ is rational and an algebraic integer, so is an integer. As $|\sin x|\le1$ then $\sin x\in\{-1,-\frac12,0,\frac12,1\}$.

Alas then, the only solutions to your problem are the obvious ones...

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  • $\begingroup$ Rather than x being a rational multiple of $\pi$, I want x to be a rational number itself when expressed in terms of degrees. $\endgroup$ – kaushal agrawal Aug 21 '17 at 3:31
  • $\begingroup$ @kaushalagrawal It's the same thing. $\endgroup$ – Lord Shark the Unknown Aug 21 '17 at 3:36
  • $\begingroup$ Oh right. Thanks $\endgroup$ – kaushal agrawal Aug 21 '17 at 4:23

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