A) \[n\pi +\frac{\pi }{4}\]
B) \[2n\,\,\pi \pm \frac{\pi }{4}\]
C) \[n\,\,\pi \pm \frac{\pi }{4}\]
D) \[n\,\,\pi -\frac{\pi }{4}\]
Correct Answer: B
Solution :
Given, \[|\sin x|=\cos x\] \[\therefore \] \[{{\sin }^{2}}x={{\cos }^{2}}x\] \[\Rightarrow \] \[1-{{\cos }^{2}}x={{\cos }^{2}}x\] \[\Rightarrow \] \[2{{\cos }^{2}}x=1\] \[\Rightarrow \] \[\cos x=\pm \frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[\cos \,x=+\frac{1}{\sqrt{2}}\] (\[\because \] cos x cannot be negative) \[\Rightarrow \] \[x=2n\,\pi \pm \frac{\pi }{4}\]
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