A) \[\alpha /2\]
B) \[\alpha \]
C) \[2\alpha \]
D) \[\alpha /6\]
Correct Answer: A
Solution :
\[\sin \theta +\sin \,3\theta +\sin \,2\theta =\sin \,\alpha \] Þ \[2\sin 2\theta \cos \theta +\sin 2\theta =\sin \alpha \] Þ \[\sin 2\theta (2\cos \theta +1)=\sin \alpha \] ?..(i) Now \[\cos \theta +\cos 3\theta +\cos 2\theta =\cos \alpha \] \[2\cos 2\,\theta \cos \,\theta +\cos 2\theta =\cos \alpha \] \[\cos 2\theta \,(2\cos \theta +1)=\cos \alpha \] ?..(ii) From (i) and (ii), \[\tan 2\theta =\tan \alpha \] Þ \[2\theta =\alpha \] Þ \[\theta =\alpha /2\].
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