<s>[INST] $\sin3x-\sin2x-\sin x=0$ 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.
 [/INST] $$\sin { 3x-\sin { 2x } -\sin { x } =0 } \\ \left( \sin { 3x-\sin { x }  }  \right) -\sin { 2x } =0\\ 2\sin { \frac { 3x-x }{ 2 } \cos { \frac { 3x+x }{ 2 }  } -\sin { 2x }  } =0\\ 2\sin { x\cos { 2x-2\sin { x\cos { x }  } =0 }  } \\ \sin { x } \left( \cos { 2x-\cos { x }  }  \right) =0\\ \sin { x } =0,\cos { 2x-\cos { x } =0 } \Rightarrow 2\cos ^{ 2 }{ x } -\cos { x } -1=0\\ $$
Can you finish?
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