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