Trigonometric Identities Question 276
Question: The least difference between the roots, in the first quadrant $ ( 0\le x\le \frac{\pi }{2} ), $ of the equation $ 4\cos x(2-3{{\sin }^{2}}x)+(\cos 2x+1)=0 $ is
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{2} $
Show Answer
Answer:
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 $
BETA
