$\sin3x-\sin2x-\sin x=0$ I have issue solving this equation.
So I wrote $$\sin3x= 3\sin x-4\sin^3x$$ and $$\sin2x = 2\sin x\cos x$$
So we have $$3\sin x-4\sin^3x-(2\sin x\cos x)-\sin x = 0$$
But now I have $$\sin x$$ and $$\cos x$$ as unknown, and I don't know how to finish this.
 A: Hint: You can rearrange what you have to get
$$\sin x(1-\cos x)-2\sin^3 x=0\,.$$
So either $\sin x = 0$ or $1-\cos x-2\sin^2x=0$. 
For the second case, use the fact that $\sin^2x+\cos^2x=1$.
A: $$\sin { 3x-\sin { 2x } -\sin { x } =0 } \\ \left( \sin { 3x-\sin { x }  }  \right) -\sin { 2x } =0\\ 2\sin { \frac { 3x-x }{ 2 } \cos { \frac { 3x+x }{ 2 }  } -\sin { 2x }  } =0\\ 2\sin { x\cos { 2x-2\sin { x\cos { x }  } =0 }  } \\ \sin { x } \left( \cos { 2x-\cos { x }  }  \right) =0\\ \sin { x } =0,\cos { 2x-\cos { x } =0 } \Rightarrow 2\cos ^{ 2 }{ x } -\cos { x } -1=0\\ $$
Can you finish?
A: it can be factorized into
$$-8\cos(x/2)\sin(x/2)^3(1+2\cos(x))=0$$
A: As the arcs are in arithmetic progression, rewrite the equation as
\begin{align}
\sin(2x+x)-\sin 2x -\sin(2x-x)&=2\sin x\cos 2x-\sin 2x=2\sin x\cos 2x-2\sin x\cos x \\
&=2\sin x(\cos2x-\cos x)
\end{align}
So one has to solve:


*

*$\sin x=\iff x\equiv 0\mod\pi$.

*$\cos 2x=\cos x \iff 2x\equiv\pm x\mod 2\pi\iff\begin{cases}x\equiv 0\mod 2\pi\quad\text{(included in  the solutions of the previous equation)}\\
3x\equiv 0\mod2\pi\iff x\equiv 0\mod\frac{2\pi}3\quad\text{(contains the previous solutions)}\end{cases}$

