Question

If $4( \sin \; 2x \; \sin \; 4x + \sin^2 x) = 3$ then $x =$

• A. $\frac{2 n \pi}{3} \pm \frac{\pi}{9} , n \in Z $
• B. $\frac{n \pi}{3} \pm \frac{\pi}{9} , n \in Z$
• C. $ \frac{n \pi}{3} + (-1)^n \frac{\pi}{9} , n \in Z $
• D. $\frac{n \pi}{3} + ( -1)^n \frac{2 \pi}{9} , n \in Z $

Answer 1

Answer: (B)

Given, $4\left(\sin 2 x \sin 4 x+\sin ^{2} x\right)=3$
$\Rightarrow 2(\cos 2 x-\cos 6 x)+2(1-\cos 2 x)=3 $
$\Rightarrow 2 \cos 2 x-2 \cos 6 x+2-2 \cos 2 x=3 $
$ \Rightarrow -2 \cos 6 x+2=3 $
$ \Rightarrow -2 \cos 6 x=1$
$ \Rightarrow \cos 6 x=-\frac{1}{2} $
$ \therefore 6 x=2 n \pi \pm \frac{2 \pi}{3} $
$ \Rightarrow x=\frac{2 n \pi}{6} \pm \frac{2}{6} \times \frac{\pi}{3}$
$=\frac{n \pi}{3} \pm \frac{\pi}{9}$