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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.
Notice that: $$\sin\left(\alpha+\frac\pi{3}\right) = \sin\left(\alpha\right)\cos\left(\frac\pi{3}\right) +\cos\left(\alpha\right)\sin\left(\frac\pi{3}\right) = \\ = \frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha).$$
Moreover: $$\sin\left(2\alpha+\frac\pi{3}\right) = \sin\left(\alpha+\left(\alpha+\frac\pi{3}\right)\right) = \sin\left(\alpha\right)\cos\left(\alpha+\frac\pi{3}\right) +\cos\left(\alpha\right)\sin\left(\alpha+\frac\pi{3}\right).$$
Recall that $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta).$ Then, the previous become:
$$\sin(\alpha)\left(\cos(\alpha)\cos\left(\frac{\pi}{3}\right) - \sin(\alpha)\sin\left(\frac{\pi}{3}\right)\right) + \cos(\alpha)\left(\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha)\right) = \\ =\sin(\alpha)\left(\frac{1}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\sin(\alpha)\right) + \cos(\alpha)\left(\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha)\right) = \\ =\sin(\alpha)\cos(\alpha)-\frac{\sqrt{3}}{2}\sin^2(\alpha) + \frac{\sqrt{3}}{2}\cos^2(\alpha).$$
Let's join together these results!
$$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0 \Rightarrow \\ \frac{483}{2}\sin(\alpha) + \frac{483\sqrt{3}}{2}\cos(\alpha) + 16\sqrt{3}\sin(\alpha)\cos(\alpha) + 24(\cos^2(\alpha) - \sin^2(\alpha)) + 20 = 0 \Rightarrow \\ \frac{483}{2}\sin(\alpha) + \frac{483\sqrt{3}}{2}\cos(\alpha) + 16\sqrt{3}\sin(\alpha)\cos(\alpha) + 48\cos^2(\alpha) - 4 = 0. $$
The last equation can be solved by setting $X = \cos(\alpha)$ and $Y = \sin(\alpha)$ with the equation $X^2 + Y^2 = 1.$:
$$\begin{cases} \frac{483}{2}Y + \frac{483\sqrt{3}}{2}X + 16\sqrt{3}XY + 48X^2 - 4 = 0 \\ X^2 + Y^2 = 1 \end{cases}.$$
Anyway, the last system of equations is a cute beast, very hard to be solved.
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 $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$ $$\alpha\sim\frac 23 \pi+\frac{4}{16 \sqrt{3}-483}$$
