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Let $n$ be a positive integer. Can we precisely solve the equation $$\sin(x) = n\cos(x)$$ in $x$?

For $n=1$, we get $x=\pi/4$.

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To add to the answer by @Fredrik, note that you don't lose any solutions by dividing by $\cos$ since both sides cannot be zero simultaneously. – S.D. Dec 1 '11 at 11:30
The set of solutions is $x_n+\pi\mathbb Z$, for some $x_n$ in $[\pi/4,\pi/2)$ depending on $n\geqslant1$, such that $x_1=\pi/4$, $x_n\lt x_{n+1}$ for every $n$ and $x_n=\pi/2-1/n+o(1/n)$ when $n\to\infty$. – Did Dec 1 '11 at 13:23
By the way: 18 minutes. – Did Dec 1 '11 at 13:26
up vote 1 down vote accepted

The only rational values of $\tan(x)$ or $\cot(x)$ for rational $x$ in degrees are $0$ and $\pm 1$. See for instance

Olmsted, J. M. H.; Discussions and Notes: Rational Values of Trigonometric Functions. Amer. Math. Monthly 52 (1945), no. 9, 507–508.

From this you cannot expect that your equation has any other nice solutions.

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Related to this result is Niven's theorem. – J. M. Dec 1 '11 at 11:59

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