Question

The number of values of $\theta \in (-\pi ,\pi ),$ satisfying $\sin 5\theta \cos 3\theta =\sin 6\theta \cos 2\theta ,$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Given, $\sin 5\theta \cos 3\theta =\sin 6\theta \cos 2\theta$
$\Rightarrow$ $2\sin 5\theta \cos 3\theta =2\sin 6\theta \cos 2\theta$
$\Rightarrow$ $\sin 8\theta +\sin 2\theta =\sin 8\theta +\sin 4\theta$ $[\because 2\sin A\cos B=\sin (A+B)+\sin (A-B)]$
$\Rightarrow$ $\sin 4\theta -\sin 2\theta =0$
$\Rightarrow$ $2\cos \frac{6\theta }{2}\sin \frac{2\theta }{2}=0$ $\left[\because \sin C-\sin D=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2}\right]$
$\Rightarrow$ $2\cos 3\theta \sin \theta =0$
$\Rightarrow$ $\cos 3\theta =0$ or $\sin \theta =0$
$\Rightarrow$ $\cos 3\theta =\cos \frac{\pi }{2}$ or $\sin \theta =\sin 0$
$\Rightarrow$ $3\theta =2n\pi \pm \frac{\pi }{2}$ or $\theta =n\pi$
$\Rightarrow$ $\theta =n\left(\frac{2\pi }{3}\right)\pm \frac{\pi }{6}$ or $\theta =n\pi$ Since, $\theta \in (-\pi ,\pi )$ So, values of $\theta$ in $(-\pi ,\pi )$ are $-\frac{\pi }{6},0,\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2}$