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I am looking for a proof, a hint or an idea to the following problem:

Is the unique solution $x\in (0,2\pi)$ of

$$ x\sin(x) + \cos(x) = 1 $$

which is equivalent to

$$ 2\arctan(x) = x$$

a rational multiple of $\pi$. I.e. is $\frac{x}{\pi} \in \mathbb{Q}$?

I believe that this is not true. This idea is based on the numeric solution, which does not look very rational:

Wolfram Alpha Solve: $x\pi\sin(x\pi) + \cos(x\pi) = 1, x\in(0,1)$ or Wolfram Alpha Solve: $2\arctan(x\pi) = \pi x, x\in(0,1)$

One idea is to use Thomas Andrews answer:

$\arctan(x)$ is a rational multiple of $\pi$ if and only if the complex number $1+xi$ has the property that $(1+xi)^n$ is a real number for some positive integer $n$. This is not possible if $x$ is a rational, $|x|\neq 1$, because $(q+pi)^n$ cannot be real for any $n$ if $(q,p)=1$ and $|qp|>1$. So $\arctan(\frac{p}{q})$ cannot be a rational multiple of $\pi$. (His full answer and proof can be found here: ArcTan(2) a rational multiple of $\pi$?)

Now, one would need to show: If $x$ is a rational multiple of $\pi$, is there an $n$, such that $(q+p\pi i)^n$ is real? For $p,q,n \in \{1,\ldots,100\}$ Mathematica says no:

Do[Do[Do[If[(p + Pi*I*q)^n  \[Element]  Rationals, 
    Print[n, p, q],], {n, 1, 100}], {p, 1, 100}], {q, 1, 100}]

Thanks in advance.

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Your equation is equivqlent to $$\tan \frac{x}{2} =x$$ bu since for $n\in \mathbb{N} $ the number $\tan\frac{\pi}{n} $ is algebraic since it is a solution of equation $\tan\left(n\cdot \frac{\pi}{n} \right) =0$ and the funcion $\tan (nx)$ is a rational funcion of $\tan x.$ tThus $\tan\frac{p\pi}{q} $ is an algebraic number for $p,q\in \mathbb{Z} . $ Hence the solution of your equation can not be of the form $\frac{p\pi}{q}.$

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  • $\begingroup$ Why is $f(x)\mapsto f(nx)$ a rational function for $f=\tan,\,n\in\mathbb N$? $\endgroup$ – dbanet Sep 17 '16 at 11:56
  • $\begingroup$ There is a recursion formula, see for example here mathworld.wolfram.com/Multiple-AngleFormulas.html $\tan(nx)=\frac{\tan((n-1)x)+\tan(x)}{1-\tan((n-1)x)\tan(x)}$ $\endgroup$ – DMNDS Sep 17 '16 at 12:04

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