A) \[n\pi \pm \frac{\pi }{6}\]
B) \[n\pi \pm \frac{\pi }{3}\]
C) \[n\pi \pm \frac{\pi }{4}\]
D) \[n\pi \pm \frac{\pi }{2}\]
Correct Answer: B
Solution :
\[3\sin \alpha -4{{\sin }^{3}}\alpha =4\sin \alpha ({{\sin }^{2}}x-{{\sin }^{2}}\alpha )\] \[\therefore \] \[{{\sin }^{2}}x={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}\] Þ \[{{\sin }^{2}}x={{\sin }^{2}}\pi /3\] Þ \[x=n\pi \pm \pi /3\].
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