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} $