<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] 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 $x$
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