Question

The sum of all the solutions of the equation $\cos x\cdot \cos \left(\frac{\pi }{3}+x\right)\cdot \cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}$ , $x\in [0,6\pi ]$ is

• A. $15\pi$
• B. $30\pi$
• C. $\frac{110\pi }{3}$
• D. None of these

Answer 1

Answer: (B)

We have, $\cos x\cdot \cos \left(\frac{\pi }{3}+x\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}$
$\Rightarrow \cos x\left(\frac{1}{4}{\cos }^{2}x-\frac{3}{4}{\sin }^{2}x\right)=\frac{1}{4}$
or $\frac{\cos x}{4}\left(4{\cos }^{2}x-3\right)=\frac{1}{4}$
$\Rightarrow 4{\cos }^{3}x-3\cos x=1$
$\left(\because \cos 3x=4{\cos }^{3}x-3\cos x\right)$
or $\cos 3x=1$
$\Rightarrow 3x=2n\pi$
$\Rightarrow x=\frac{2n\pi }{3}$, where $n=0,1,2,3,4,5,6,7,8,9$
$(\because x\in [0,6\pi ])$
$\therefore$ The required sum $=\frac{2\pi }{3}\sum _{n=0}^{9}n=30\pi$