algebra precalculus - real solution of eqn. in $sin x+2sin 2x-sin 3x = 3,$ where $xin (0,pi)$.

The no. of real solution of the equation $\sin x+2\sin 2x-\sin 3x = 3,$ where $x\in (0,\pi)$.

$\bf{My\; Try::}$ Given $\left(\sin x-\sin 3x\right)+2\sin 2x = 3$

$\Rightarrow -2\cos 2x\cdot \sin x+2\sin 2x = 3\Rightarrow -2\cos 2x\cdot \sin x+4\sin x\cdot \cos x = 3$

$\Rightarrow 2\sin x\cdot \left(-\cos 2x+2\cos x\right)=3$

Now I did not understand how can i solve it.

Help me

Thanks

Answer

$$\begin{cases}\sin 2x=2\sin x\cos x\\{}\\\sin 3x=\sin2x\cos x+\sin x\cos2x=2\sin x\cos^2x+\sin x(1-2\sin^2x)\end{cases}$$

Thus we get

$$0=\sin x+4\sin x\cos x-2\sin x\cos^2x-\sin x+2\sin^3x$$

Divide al through by $\;\sin x\;$ (why can we?):

$$4\cos x-2\cos^2x+2\sin^2x=0\iff2\cos x-\cos^2x+1-\cos^2x=0\iff$$

$$2\cos^2x-2\cos x-1=0\iff \ldots$$