A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
E) None of these
Correct Answer: E
Solution :
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 \[-\frac{\pi }{6},0,\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2}\]
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