Question

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

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

Answer 1

The correct option is B $30\pi$

$\cos \theta \cos (\frac{\pi }{3}+\theta )\cos (\frac{\pi }{3}-\theta )=\frac{1}{4}$
$\Rightarrow 4\cos \theta \cos (\frac{\pi }{3}+\theta )\cos (\frac{\pi }{3}-\theta )=1$
$\Rightarrow 2\cos \theta (\cos \frac{2\pi }{3}+\cos 2\theta )=1$
$\Rightarrow 4{\cos }^3\theta -3\cos \theta =1$
$\Rightarrow \cos 3\theta =1$
$\Rightarrow 3\theta =2n\pi ,n\in Z$
$\Rightarrow \theta =\frac{2n\pi }{3}$

Possible values of $n$ is $0,1,2,\cdots ,9$
Required sum $=\frac{2\pi }{3}\sum _{n=1}^{9}n=30\pi$