<s>[INST] real solution of eqn. in $\sin x+2\sin 2x-\sin 3x = 3,$ where $x\in (0,\pi)$. 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
 [/INST] 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 $\sin x+2\sin 2x-\sin 3x =3\;\forall x\in (0,\pi)$
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