A) \[\pm \frac{\pi }{6}\]
B) \[\pm \frac{\pi }{4}\]
C) \[\pm \frac{3\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
We have \[1-\cos 2x+1-{{\cos }^{2}}2x=2\] or \[\cos 2x(\cos 2x+1)=0\] \[\therefore \,\,\cos 2x=0,\,-1,\] \[\therefore \,\,2x=\left( n+\frac{1}{2} \right)\pi \] or \[(2n+1)\,\pi \] \[\Rightarrow \,\,\,x=(2n+1)\frac{\pi }{4}\] or \[(2n+1)\frac{\pi }{2}\] Now, put \[n=-2,-1,0,1,21\] \[\therefore \,\,x=\frac{-3\pi }{4},\frac{-\pi }{4},\frac{\pi }{4},\frac{5\pi }{4}\], and \[\frac{-3\pi }{2},\frac{-\pi }{2},\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\] Since \[-\pi \le x\le \pi ,\]therefore, \[x=\pm \frac{\pi }{4},\pm \frac{3\pi }{4}\] only.
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