A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{12}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: C
Solution :
\[\sin 5x+\sin 3x+\sin x=0\] \[\Rightarrow \] \[-\sin 3x=\sin 5x+\sin x=2\sin 3x\cos 2x\] \[\Rightarrow \] \[\sin 3x=0\] \[\Rightarrow \] \[x=0\] or \[\cos 2x=-\frac{1}{2}=-\cos \,\left( \frac{\pi }{3} \right)=\cos \,\left( \pi -\frac{\pi }{3} \right)\] \[\Rightarrow \] \[2x=2n\pi \pm \left( \pi -\frac{\pi }{3} \right)\,\Rightarrow x=n\pi \pm \left( \frac{\pi }{3} \right)\] For x lying between 0 and \[\frac{\pi }{2}\], we get \[x=\frac{\pi }{3}\]. Trick: Check with options.
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