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} $