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I'm experimenting with functions and several operators and stumbled across this problem:

$x \sin(n) = n \sin(x)$

What are the solutions of x for each $n \in \mathbb{Z}$? I tried to input this on WolframAlpha but no close formula for the solution were found, It might hide something like $2\pi n + \arcsin(y), n \in \mathbb{Z}$ for the solutions of $y = \sin(x)$. Converting $\sin(x)$ to its complex form might be the case but I personally didn't gain much from it. Can an expert user help me out in this problem? Thanks.

Kinesis
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    Of course $x=n$ is a solution. I doubt that there is a closed form for the other solutions except in the trivial case where $\sin(n) = 0$. – Martin R Nov 15 '21 at 14:35
  • So essentially you want an inverse of $\operatorname{sinc}$... Well certainly $\operatorname{sinc}$ isn't periodic, therefore you have no hope of $2\pi k+\text{stuff}$ outside of $n=0$. –  Nov 15 '21 at 14:36
  • Related: https://math.stackexchange.com/q/2175174/42969 – Martin R Nov 15 '21 at 14:37
  • @Saucy O'Path Well yes, mine was just an example of how I need the solutions to look like in order to continue with my task – Kinesis Nov 15 '21 at 14:42
  • @pH 74 still that's not a closed form of the solutions, and I highly doubt most x solutions will be integers lol. – Kinesis Nov 15 '21 at 14:45
  • Using the bound $|\sin(x)/x|\le 1/x$ for any fixed value $y_0=\sin(n)/n$ one can see that there will be finitely many solutions which will lie in $[-1/y_0,1/y_0]$. Since $\sin(x)/x$ is even if $x_0$ is a solution, then so is $-x_0$. Plotting the function we can see geometrically from it's periodicity that there will be two solutions in the interval $[2\pi k,2\pi(k+1)]$ for each integer $1\le k\le 1/y_0$ (as long as $y_0\neq 0$) together with either 1 or two solutions in the interval $[0,2\pi]$ depending on whether $y_0$ is positive or negative. One could then approximate these solutions. – David Sheard Nov 15 '21 at 14:54
  • For $k\ge 3$ or so, because $1/x$ is quite flat, I think $\sin(x)/x$ can be closely approximated in the interval $[2\pi k, 2\pi (k+1)]$ by the scaled $\sin$ function, $\sin(x)/2\pi k$. Therefore you could get good approximate solutions using normal trig, and give a closed form set of approximate solutions which look a bit like the form you wrote in your question. If you really need exact solutions, I'm afraid I can't see a straightforward way to find those. – David Sheard Nov 15 '21 at 15:01

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