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
$$\begin{cases}\sin 2x=2\sin x\cos x\\{}\\\sin 3x=\sin2x\cos x+\sin x\cos2x=2\sin x\cos^2x+\sin x(1-2\sin^2x)\end{cases}$$
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?):
$$4\cos x-2\cos^2x+2\sin^2x=0\iff2\cos x-\cos^2x+1-\cos^2x=0\iff$$
$$2\cos^2x-2\cos x-1=0\iff \ldots$$
Given $\sin x+2\sin 2x-\sin 3x = 3\Rightarrow (\sin 2x-1)^2+(\cos 2x+\sin x)^2+(\cos x)^2=0\; \forall x\in (0,\pi)$
So either $\sin 2x = 1$ and $\cos 2x = -\sin x$ and $\cos x=0$
So from above no common value of $x\in (0,\pi)$ for which above equation is satisfied.
So no real values of $x$
Using Inequality
Starting from $\bf{L.H.S\;\;}$
$$\sin x+2\sin 2x-\sin 3x = -2\sin x\cos 2x+2\sin 2x\leq \left|2\sin x\cos 2x+2\sin 2x\right|$$
$$\leq 2|\sin 2x+\cos 2x|\leq 2\sqrt{2}\left|\sin \left(2x+\frac{\pi}{4}\right)\right|\leq 2\sqrt{2}<3\;(\bf{R.H.S})$$
So no real values of $x$ for which $\sin x+2\sin 2x-\sin 3x =3\;\forall x\in (0,\pi)$