Question

The equation $\sin 3\theta =4\sin \theta \cdot \sin 2\theta .\sin 4\theta$ 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

Answer: (D)

$\sin 3\theta =4\sin \theta \sin 2\theta \sin 4\theta$
$\Rightarrow \sin 3\theta =(2\sin \theta )(2\sin 2\theta \sin 4\theta )$
$\Rightarrow 3\sin \theta -4{\sin }^3\theta =2\sin \theta (\cos 2\theta -\cos 6\theta )$
$\Rightarrow 3-4{\sin }^2\theta =2(\cos 2\theta -\cos 6\theta )$ or $\sin \theta =0$
$\Rightarrow 3-2(1-\cos 2\theta )=2\cos 2\theta -2\cos 6\theta$ or $\sin \theta =0$
$\Rightarrow 1=-2\cos 6\theta \Rightarrow \cos 6\theta =\frac{-1}{2}$ or $\sin \theta =0$
$\therefore \sin \theta =0$ or $\cos 6\theta =\frac{-1}{2}$
$\Rightarrow \theta =n\pi \text{ or }\theta =\frac{2n\pi \pm \left(\frac{2\pi }{3}\right)}{6}=\frac{n\pi }{3}\mp \frac{\pi }{9}$
$\Rightarrow \theta =0,\pi ,\frac{\pi }{9},\frac{\pi }{3}\mp \frac{\pi }{9},\frac{2\pi }{3}\mp \frac{\pi }{9},\pi -\frac{\pi }{9}$
So eight solutions.