Question

If $\cos \, 3x \, \cos \, 2x \, \cos \, x = \frac{1}{4}$ and $0 < x < \frac{\pi}{4}$, then the value of $x$ is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{5}$
• C. $\frac{\pi}{8}$
• D. $\frac{\pi}{7}$

Answer 1

Answer: (C)

$\frac{\pi}{8}$

Answer 2

Given, $\cos\,3x\, \cos\,2x\, \cos\, x=\frac{1}{4}$
$\Rightarrow 2\left(2\cos\, x\, \cos\,3x\right)\, \cos\, 2x=1$
$\Rightarrow 2\left(\cos\,4x + \cos\, 2x\right)\, \cos\, 2x=1$
$\left[\because 2\cos\, A\, \cos\, B-\cos\, B=\cos\left(A+B\right) + \cos\left(A-B\right)\right]$
$\Rightarrow 2\left(2\cos^{2}\, 2x-1 + \cos\, 2x\right)\, \cos\, 2x=1$
$\Rightarrow 4\, \cos^{3}\, 2x-2\cos\, 2x+2\cos^{2}\, 2x-1=0$
$\Rightarrow 2\cos\, 2x\left(2\cos^{2}\, 2x-1+1\left(2\cos^{2}\,x-1\right)\right)=0$
$\Rightarrow \left(2\cos^{2}\, 2x-1\right)\left(2\cos\,2x+1\right)=0$
$\Rightarrow \cos\,4x(2\cos\,2x + 1) = 0$
$\Rightarrow \cos\,4x=0$ or $2\cos\,2x + 1=0$
$\Rightarrow \cos\,4x =\cos \frac{\pi}{2}$ or $\cos\,2x=\frac{-1}{2}$
$\Rightarrow 4x=\frac{\pi}{2}$ or $\cos\,2x=\cos \frac{2\pi}{3}$
$\Rightarrow x=\frac{\pi}{8}$ or $2x=\frac{2\pi}{3}$
$\Rightarrow x=\frac{\pi}{8}$ or $x=\frac{\pi}{3}$
Since, $x \in\left(0, \frac{\pi}{4}\right)$
$\therefore x=\frac{\pi}{8}$ is the required value.