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

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

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

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

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

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

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

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 $n\geq3$?