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)

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