<s>[INST] Number of solutions of this trigonometric equation. 
Q. Find the number of solutions of the equation $\sin(x) + 2\sin(2x) - \sin(3x) = 3$, in the interval $x\in (0,\pi)$.  

I tried clubbing the $\sin(x)$ and $\sin(3x)$ terms together but got nothing. I also tried the $\sin(x)$ with $\sin(2x)$ and $\sin(2x)$ with $\sin(3x)$. How do i do it?
 [/INST] We have
\begin{align}
2\sin 2x+\sin x-\sin 3x &=4\sin x\cos x-2\sin x \cos 2x=\\
&=\sin x(4\cos x -4\cos^2 x+2)=\\
&=\sin x  \left[3-(2\cos x-1)^2\right].
\end{align}
Since $\sin x\ge 0$ in the interval $[0,\pi]$ and the second factor is bounded above by $3$, we must have simultaneously $\sin x=1$, $2\cos x=1$, which is not possible. Therefore the equation has no solutions in $[0,\pi]$.
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