A) \[n\pi +{{(-1)}^{n}}\frac{\pi }{2},\,n\in Z\]
B) \[\frac{n\pi }{2},\,n\in Z\]
C) \[(4n\pm 1)\frac{\pi }{2},\,n\in Z\]
D) \[(2n-1)\pi ,\,n\in Z\]
Correct Answer: C
Solution :
\[{{\sin }^{2}}x-\cos 2x=2-\sin 2x\] \[\Rightarrow \]\[1-{{\cos }^{2}}x-(2{{\cos }^{2}}x-1)\] \[=2-2\sin x\cos x\] \[\Rightarrow \]\[-3{{\cos }^{2}}x+2\sin x\cos x=0\] \[\Rightarrow \]\[\cos x(2\sin x-3\cos x)=0\] \[\Rightarrow \]\[\cos x=0,\] \[(\because \,2\sin x-3\cos x\ne 0)\] \[\Rightarrow \] \[x=2n\pi \pm \frac{\pi }{2}\] \[\Rightarrow \] \[x=(4n\pm 1)\frac{\pi }{2},\,n\in Z\]
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