A) \[n\,\pi \] or \[2n\,\pi +\frac{\pi }{2}\]
B) \[n\,\pi \]or \[2n\,\,\pi +\frac{\pi }{2}\] or \[n\,\,\pi +{{(-1)}^{n}}\frac{\pi }{6},\] \[n\in I\]
C) \[n\,\,\pi +{{(-1)}^{n}}\frac{\pi }{6},\,n\in I\]
D) None of these
Correct Answer: C
Solution :
The equation holds if \[|\cos x|=1\] i.e., if \[x=n\pi ,\,\,n\in I\] If \[|\cos x|\ne 1\] then \[{{\sin }^{2}}x-\frac{3}{2}\sin x+\frac{1}{2}=0\] \[\Rightarrow \,\,\sin x=1\] or \[\frac{1}{2}\] \[\sin x\ne 1,\]as in that case \[cos\text{ }x=0\] \[\therefore \,\,\sin x=\frac{1}{2}\Rightarrow \,\,x=n\pi +{{(-1)}^{n}}\frac{\pi }{6}\]
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