A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
We have, \[4\cos x(2-3{{\sin }^{2}}x)+(\cos 2x+1)=0\] \[\Rightarrow \,\,4\cos x(3{{\cos }^{2}}x-1)+2{{\cos }^{2}}x=0\] \[\Rightarrow \,\,2\cos x(6{{\cos }^{2}}x+\cos x-2)=0\] \[\Rightarrow \,\,2\cos x(3\cos x+2)(2cosx-1)=0\] \[\Rightarrow \]either \[\cos \,\,x=0\] which gives \[x=\pi /2\] or \[\cos \,x=-2/3\] Which gives no value of x for which \[0\le x\le \pi /2\] or \[cos\,x=1/2,\]which gives \[x=\pi /3\] So, the required difference \[=\pi /2-\pi /3=\pi /6\]You need to login to perform this action.
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