How to solve the equation $\sin^3x+\sin^3(\frac{2\pi}{3}+x)+\sin^3(\frac{4\pi}{3}+x)+\frac{3}{4}\cos2x=0$. Solve the equation $$\sin^3x+\sin^3\left(\frac{2\pi}{3}+x\right)+\sin^3\left(\frac{4\pi}{3}+x\right)+\frac{3}{4}\cos2x=0$$
May I have a hint on how to solve this equation?
 A: Using the identity $\sin^3 \theta = \dfrac{3}{4}\sin\theta-\dfrac{1}{4}\sin 3\theta$, we get: 
$\sin^3 x + \sin^3(x+\tfrac{2\pi}{3}) + \sin^3(x+\tfrac{4\pi}{3})$ 
$= \dfrac{3}{4}\left[\sin x + \sin(x+\tfrac{2\pi}{3})+\sin(x+\tfrac{4\pi}{3})\right] - \dfrac{1}{4}\left[\sin 3x + \sin(3x+2\pi) + \sin(3x+4\pi)\right]$
$= \dfrac{3}{4} \cdot 0 - \dfrac{1}{4} \cdot 3\sin3x$
$= -\dfrac{3}{4}\sin 3x$. 
Therefore, your equation is equivalant to $-\dfrac{3}{4}\sin 3x + \dfrac{3}{4}\cos 2x = 0$. 
Can you take it from here? (The next thing you should do is use one of these sum-to-product identities.)

Using the identities $\sin\theta = \cos(\tfrac{\pi}{2}-\theta)$ and $\cos A - \cos B = -2\sin\tfrac{A+B}{2}\sin\tfrac{A-B}{2}$, we have:
$\cos 2x - \sin 3x= 0$
$\cos 2x -\cos(\tfrac{\pi}{2}-3x) = 0$
$2\sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right) = 0$
$\sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right) = 0$ OR $\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right) = 0$
$\dfrac{x}{2}-\dfrac{\pi}{4} = k\pi$ OR $\dfrac{5x}{2}-\dfrac{\pi}{4} = k\pi$ for some $k \in \mathbb{Z}$
$x = \dfrac{(4k+1)\pi}{2}$ OR $x = \dfrac{(4k+1)\pi}{10}$ for some $k \in \mathbb{Z}$
It's not hard to see that $x = \dfrac{(4k+1)\pi}{10}$ covers all the values which satisfy $x = \dfrac{(4k+1)\pi}{2}$. So the solution is simply $x = \dfrac{(4k+1)\pi}{10}$ for some $k \in \mathbb{Z}$.
