Question

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

• A. $15\pi$
• B. $30\pi$
• C. $\frac{100\pi }{3}$
• D. None of these

Answer 1

Answer: (B)

$2\cos \theta \left(\cos 120^{\circ }+\cos 2\theta \right)=1$
$\Rightarrow 2\cos \theta \left(-\frac{1}{2}+2{\cos }^2\theta -1\right)=1$
$\Rightarrow 4{\cos }^3\theta -3\cos \theta -1=0$
$\Rightarrow \cos 3\theta =1=\cos 0^{\circ }$
$\Rightarrow 3\theta =2n\pi$ or $\theta =\frac{2n\pi }{3}$
Give values so that $2n$ does not exceed $18$
$\therefore n=0,1,2,3,\dots ,9$
Hence, the sum $=\frac{6\pi }{18}\sum _0^{9}2n=\frac{2\pi }{3}\sum _1^{9}n$
$=\frac{2\pi }{3}\frac{9(9+1)}{2}=30\pi$