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 $2 \cos 2\, \theta+1=0$
$\Rightarrow 4 \,\theta=(2 n+1) \frac{\pi}{2}, \cos 2 \,\theta=-\frac{1}{2}$
$\Rightarrow \theta=(2 n+1) \frac{\pi}{8}, 2 \,\theta=2 \pi / 3$ 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$ .