Answer:
Option D
Explanation:
Plan for solving this type of questions , obtain the LHS and RHS in equation and examine , the two are equal or not for a given interval
Given, Trigonometrical Equation
(sin x -sin 3x)+2 sin2x =3
$\Rightarrow$ $ -2\cos 2x\sin x+4 \sin x \cos x=3$
$[\because $ $ \sin C-\sin D=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) $ and $\sin2\theta=2\sin\theta\cos\theta]$
$\Rightarrow $ $2\sin x(2\cos x-\cos 2x)=3$
$\Rightarrow $ $ 2\sin x(2\cos x-\cos^{2} x+1)=3$
$\Rightarrow $ $2\sin x\left(\frac{3}{2}-2\left(\cos x-\frac{1}{2}\right)^{2}\right)=3$
$\Rightarrow $ $3\sin x-3=4(\cos x-\frac{1}{2})^{2}\sin x$
As $x\epsilon (0,\pi)$
$LHS \leq 0 $ and $ RHS\geq0$
for solution to exist LRS=RHS
Now, LHS =0 $\Rightarrow $ 3 sin x-3=0
$\Rightarrow $ sin x=1
$\Rightarrow $ $x=\frac{\pi}{2}$
For $x=\frac{\pi}{2}$
RHS= $4\left(\cos\frac{\pi}{2}-\frac{1}{2}\right)^{2}\sin\frac{\pi}{2}$
= $4\left(\frac{1}{4}\right)(1)=1\neq0$
$\therefore$ No solution of the equation exists
