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