A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{12}\]
C) \[\frac{\pi }{8}\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
\[2\sin x\cos x=\frac{1}{2}\] \[\Rightarrow \] \[\sin 2x=\frac{1}{2}=\sin \frac{\pi }{6}\] \[\Rightarrow \] \[2x=n\pi +{{(-1)}^{n}}\frac{\pi }{6}\] \[\Rightarrow \] \[x=\frac{n\pi }{2}+{{(-1)}^{n}}\frac{\pi }{12}\] For \[x\in \left( 0,\frac{\pi }{2} \right)\] \[x=\frac{\pi }{12}\]
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