Question

$\sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta$ in $0\le \theta \le \pi$ has:

• A. 2 real solutions
• B. 4 real solutions
• C. 6 real solutions
• D. 8 real solutions

Answer 1

The correct option is D 8 real solutions

Given equation can be written as
$3\sin \theta -4{\sin }^3\theta =4\sin \theta \sin 2\theta \sin 4\theta$
Hence either $\sin \theta =0\Rightarrow \theta =n\pi$
or $3-4{\sin }^2\theta =4\sin 2\theta \sin 4\theta$
$3-2(1-\cos 2\theta )=2(\cos 2\theta -\cos 6\theta )$
or $1=-2\cos 6\theta$
$\cos 6\theta =-\frac{1}{2}=\cos \frac{2\pi }{3}$
$6\theta =2n\pi \pm \frac{2\pi }{3}$
if $-\pi \le \theta \le \pi$ then total solution are $0,\frac{\pi }{9},\frac{2\pi }{9},\frac{4\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9},\frac{8\pi }{9},\pi$ is $8$ real solutions.