Is the solution to this trig function known to be algebraic or transcendental? This largest solution to this gorgeous equation is the first local extremum on a function related to the Fibonacci sequence:
$$x^2 \cdot \sin \left(\frac{2\pi}{x+1} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x} \right) \right) = (x+1)^2 \cdot \sin \left(\frac{2\pi}{x} \right) \cdot \left(3+2 \cos \left(\frac{2\pi}{x+1} \right) \right)$$
This is as simplified as I could get it. The largest solution to this equation is around $x = 2.1392.$
It appears there is no closed-form solution for this; is there any way to prove if the solution is algebraic or transcendental?
P.S. Can anyone approximate this constant to more decimal places?
ANSWERED
 A: It should be possible to show that $x$ is irrational using Theorem 7 of Trigonometric diophantine equations (On vanishing sums of roots of unity) by J. H. Conway and A. J. Jones, Acta Arithmetica 30 (1976), 229–240, although I have not carried out the full calculation.  By letting $\alpha = 2\pi/(x+1)$ and $\beta = 2\pi/x$ and using standard trig identities, we can rewrite the given equation as
$$3x^2\sin \alpha -3(x+1)^2\sin\beta - (2x+1)\sin(\alpha+\beta) + (2x^2+2x+1)\sin(\alpha-\beta) = 0.$$
We can convert from sines to cosines via $\sin \gamma \equiv \cos(\pi/2 - \gamma)$. Then Theorem 7 of Conway and Jones tells us that there are only a few "primitive" ways of getting a rational linear combination of four cosines of rational multiples of $\pi$ to vanish:
$$\eqalign{{1\over2} &= \cos{\pi\over 3}\cr
  0 &= -\cos\phi + \cos\biggl({\pi\over 3}-\phi\biggr) + \cos\biggl({\pi\over 3}+\phi\biggr)\cr
{1\over2} &= \cos{\pi\over 5} - \cos{2\pi\over 5}\cr
{1\over2} &= \cos{\pi\over 7} - \cos{2\pi\over 7} + \cos{3\pi \over 7}\cr
{1\over2} &= \cos{\pi\over 5} - \cos{\pi\over 15} + \cos{4\pi\over 15}\cr
{1\over2} &= -\cos{2\pi\over 5} + \cos{2\pi\over 15}-\cos{7\pi\over 15}\cr
{1\over2} &= \cos{\pi\over 7} + \cos{3\pi\over 7} - \cos{\pi\over 21} + \cos{8\pi\over21}\cr
{1\over2} &= \cos{\pi\over 7} -\cos{2\pi\over7}+\cos{2\pi\over21}-\cos{5\pi\over 21}\cr
{1\over2} &= -\cos{2\pi\over 7} + \cos{3\pi\over 7}+ \cos{4\pi\over21} + \cos{10\pi\over 21}\cr
{1\over2} &= -\cos{\pi\over 15}+\cos{2\pi\over15}+\cos{4\pi\over 15}-\cos{7\pi\over 15}\cr  
}$$
One should be able to go through this list and check case by case that no rational value of $x$ can yield the desired equation.
I do not know how to show that $x$ must be transcendental.
