Question

The sum of all the solution of the equations $\cos \theta \cdot \cos (\frac{\pi }{3}+\theta )\cdot \cos (\frac{\pi }{3}-\theta )=\frac{1}{4}$, where $\theta ϵ[0,6\pi ]$ is

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

Answer 1

Answer: (B)

$30\pi$

$\cos (\frac{\pi }{3}-\theta )\cos \theta \cos (\frac{\pi }{3}+\theta )=\frac{1}{4}$
$\Rightarrow ({\cos }^2\frac{\pi }{3}-{\sin }^2\theta )\cos \theta =\frac{1}{4}$
$\Rightarrow (\frac{1}{4}-{\sin }^2\theta )\cos \theta =\frac{1}{4}$
$\Rightarrow (\frac{1}{4}-1+{\cos }^2\theta )\cos \theta =\frac{1}{4}$
$\Rightarrow \frac{4{\cos }^3\theta -3\cos \theta }{4}=\frac{1}{4}\Rightarrow \frac{1}{4}\cos 3\theta =\frac{1}{4}$
$\Rightarrow \cos 3\theta =1\Rightarrow \cos 3\theta =\cos 2n\pi$
$\Rightarrow 3\theta =2n\pi$
$\Rightarrow \theta =\frac{2n\pi }{3}\Rightarrow \theta =\frac{2n}{18}\left(6\pi \right)$
$\Rightarrow 0\le \frac{2n}{18}\le 1,0\le n\le 9$
$\therefore$ Sum of all values of $\theta =\frac{6\pi }{18}\sum _{n=0}^{6\pi }2n$
$=\frac{2\pi }{3}\sum _{n=1}^{9}n=30\pi$
$\therefore$ Choice (b) is correct.