# 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.

• Are you sure the equation has a "nice" solution?
– MasB
Feb 1, 2020 at 19:03
• @BernardMassé: no,I'm not; but I would like to find it. Feb 1, 2020 at 19:04
• Maple gives a rather horrible answer when asked to solve the equation. Which is why I asked my previous question.
– MasB
Feb 1, 2020 at 19:09
• I am ready to bet that there is a typo : $20$ should be $\color{red}{24}$ Feb 2, 2020 at 10:08
• @Claude Leibovici: no, it's actually $20$. Feb 2, 2020 at 10:53

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.

• There is for sure a typo in the problem : $20$ should be $24$ and using your approach, we end with $$3 (2 \cos (2 \alpha )+1) \left(10304 \sqrt{3} \cos (\alpha )+512 \cos (2 \alpha )+78275\right)=0$$ Feb 2, 2020 at 10:20

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 $$483\sin\left(\alpha+\frac\pi{3}\right)+16\sqrt3\sin\left(2\alpha+\frac{\pi}{3}\right)+20=0$$ $$\alpha\sim\frac 23 \pi+\frac{4}{16 \sqrt{3}-483}$$