Question
The sum of all the solutions of the equation $$\cos x \cdot \cos \left( {\frac{\pi }{3} + x} \right) \cdot \cos \left( {\frac{\pi }{3} - x} \right) = \frac{1}{4},x \in \left[ {0,6\pi } \right]$$ is
A.
$${15\pi }$$
B.
$${30\pi }$$
C.
$${\frac{110\pi }{3}}$$
D.
None of these
Answer :
$${30\pi }$$
Solution :
$$\eqalign{
& {\text{Here, }}\cos x\left( {\frac{1}{4}{{\cos }^2}x - \frac{3}{4}{{\sin }^2}x} \right) = \frac{1}{4}\,\,\,\,{\text{or, }}\frac{{\cos x}}{4}\left( {4{{\cos }^2}x - 3} \right) = \frac{1}{4} \cr
& {\text{or, }}\cos 3x = 1 \cr
& \Rightarrow \,\,3x = 2n\pi \cr
& \Rightarrow \,\,x = \frac{{2n\pi }}{3},\,{\text{where }}n = 0,1,2,3,4,5,6,7,8,9. \cr} $$
∴ the required sum $$ = \frac{{2\pi }}{3}\sum\limits_{n = 0}^9 n .$$