Solving the trigonometric equation $483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$ 
I'm trying the value of $\alpha \in [0,\pi]$ that is solution to this trigonometric equation: $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$

I've tried to write down $\sin\left(\alpha+\frac\pi{3}\right)$ and $\sin\left(2\alpha+\frac{\pi}{3}\right)$with the formula $\sin(\alpha+\beta)=\sin\alpha \cos\beta+\sin\beta\cos\alpha$, but after that I'am stuck and I don't have completely any idea of how to proceed.

Starting from @the_candyman's answer
$$\begin{cases}
\frac{483}{2}Y + \frac{483\sqrt{3}}{2}X + 16\sqrt{3}XY + 48X^2 - 4 = 0 \\
X^2 + Y^2 = 1 
\end{cases}$$
eliminate $Y$ from the first equation
$$Y=\frac{-96 X^2-483 \sqrt{3} X+8}{32 \sqrt{3} X+483}$$ Plug it in the second to end with
$$12288 X^4+123648 \sqrt{3} X^3+928548 X^2-38640 \sqrt{3} X-233225=0$$ which can be solved exactely using radicals.
The formulae are messy; the usual tests for quartic equations shows only two reals roots which numerically are
$$X_1=-0.492379182948765 \qquad \text{and} \qquad X_2=+0.506762587904079$$
$$Y_1=+0.870380801832569 \qquad \text{and} \qquad Y_2=-0.862085656707476$$ 
 The solution for $\alpha \in [0,\pi]$ then corresponds to $(X_i,Y_1)$.
In fact, the solution is so close to $\frac 23 \pi$ that we could make one single Newton iteration for the original equation
$$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$
$$\alpha\sim\frac 23 \pi+\frac{4}{16 \sqrt{3}-483}$$
