For $n\geq1$, the largest solution to this lovely equation is a local extremum on a function related to the Fibonacci sequence:


$$\sum\_{k=1}^{n} k{(-1)^{k}} \cdot \frac{\sin(\frac{k\pi}{x} )}{3+2\cos(\frac{k\pi}{x} )} = 0$$


For $n=1$, the largest solution is $1$. For $n=2$, the largest solution is:


$$\frac{\pi}{2\arctan(\frac{\sqrt{4\sqrt{10}-5}}{3})} $$


Can it be proven that for any integer $n\geq2$, the solution to the above equation can be expressed as


$$\frac{\pi}{2\arctan(A)} $$


where $A$ is **algebraic**? If so, what are some minimal polynomials for $A$ when $n\geq3$?



Can the solutions to this beautiful equation always be expressed in terms of algebraic numbers?
Yes, you can write it in terms of an algebraic number as you desire. Both $\sin\frac{k\pi}{x}$ and $\cos\frac{k\pi}{x}$ can be written as polynomials in $\cos\frac{\pi}{x},\sin\frac{\pi}{x}$, which [can in turn be written](https://en.wikipedia.org/wiki/Weierstrass_substitution#The_substitution) as rational functions in $\tan\frac{\pi}{2x}$. It follows that for any solution $x$ of your equation, $A=\tan\frac{\pi}{2x}$ will be algebraic. But since $x=\frac{\pi}{2\arctan A}$, this is your desired result.


From this method you can easily figure out a polynomial of which $A$ is a root of, but I doubt there is an easy way to find its *minimal* polynomial.


