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?
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 $[0,\pi]$.
