Question

The number of solutions of the equation $4\cos 2\theta \cdot \cos 3\theta =\sec \theta$, when $0<\theta <\pi$, is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (C)

We have,
$4\cos 2\theta \cdot \cos 3\theta =\sec \theta ,($ where $0<\theta <\pi )$
$\Rightarrow 2(2\cos 2\theta \cdot \cos 3\theta )=\frac{1}{\cos \theta }$
$\Rightarrow 2[\cos 5\theta +\cos (-\theta )]=\frac{1}{\cos \theta }$
$[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)]$
$\Rightarrow 2[\cos 5\theta +\cos \theta ]=\frac{1}{\cos \theta }$
$[\because \cos (-\theta )=\cos \theta ]$
$\Rightarrow 2\cos 5\theta \cdot \cos \theta +2{\cos }^2\theta =1$
$\Rightarrow (\cos 6\theta +\cos 4\theta )+\left(2{\cos }^2\theta -1\right)=0$
$\Rightarrow \cos 6\theta +\cos 4\theta +\cos 2\theta =0$
$\left[\because 2{\cos }^2\theta -1=\cos 2\theta \right]$
$\Rightarrow 2\cos 4\theta \cdot \cos 2\theta +\cos 4\theta =0$
$\Rightarrow \cos 4\theta (2\cos 2\theta +1)=0$
$\Rightarrow \cos 4\theta =0$ and $2\cos 2\theta +1=0$
$\Rightarrow 4\theta =(2n+1)\frac{\pi }{2},\cos 2\theta =-\frac{1}{2}$
$\Rightarrow \theta =(2n+1)\frac{\pi }{8},2\theta =2\pi /3$ or $\frac{4\pi }{3}$
$\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8},\frac{\pi }{3},\frac{2\pi }{3}$
So, number of solution of given equation is $6$ .