JEE Main On 08 April 2018 Question 8
Question: If sum of all the solutions of the equation $ 8\cos x\cdot ( \cos ( \frac{\pi }{6}+x )\cdot \cos ( \frac{\pi }{6}-x )-\frac{1}{2} )=1 $ in $ [0,\pi ] $ is $ k\pi , $ then k is equal to: [JEE Main Online 08-04-2018]
Options:
A) $ \frac{8}{9} $
B) $ \frac{20}{9} $
C) $ \frac{2}{3} $
D) $ \frac{13}{9} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 8\cos x\cdot ( \cos ( \frac{\pi }{6}+x )\cdot \cos ( \frac{\pi }{6}-x )-\frac{1}{2} )=1 $
$ 8\cos x[ {{\cos }^{2}}x-{{\sin }^{2}}\frac{\pi }{6}-\frac{1}{2} ] $
$ 8\cos x[ {{\cos }^{2}}x-\frac{3}{4} ]=1 $ $ 8{{\cos }^{3}}x-6\cos x=1 $
$ {{\cos }^{3}}x=\frac{1}{2} $ $ 3x=\frac{\pi }{3},2\pi -\frac{\pi }{3},2\pi +\frac{\pi }{3} $
$ x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9} $ Sum $ =\frac{13\pi }{9} $
