Question

$\frac{\cos 3 \theta}{2 \cos 2 \theta-1}=\frac{1}{2} \text { if }$

• A. $\theta=n \pi+\frac{\pi}{3}, n \in I$
• B. $\theta=2 n \pi \pm, \frac{\pi}{3} n \in I$
• C. $\theta=2 n \pi \pm \frac{\pi}{6}, n \in I$
• D. $\theta=n \pi+\frac{\pi}{6}, n \in I$

Answer 1

Answer: (B)

$\frac{\cos 3 \theta}{2 \cos 2 \theta-1}=\frac{1}{2}$
$\Rightarrow 2\left(4 \cos ^3 \theta-3 \cos \theta\right)=2\left(2 \cos ^2 \theta-1\right)-1$
$\Rightarrow 8 \cos ^3 \theta-4 \cos ^2 \theta-6 \cos \theta+3=0$
$\Rightarrow\left(4 \cos ^2 \theta-3\right)(2 \cos \theta-1)=0$
$\Rightarrow \cos \theta=\frac{1}{2}, \pm \frac{\sqrt{3}}{2}$
But when $\cos \theta=\pm \frac{\sqrt{3}}{2}$
then $2 \cos 2 \theta-1=0$
$\therefore$ rejecting this value,
$\cos \theta=\frac{1}{2}$ is valid only
$\Rightarrow \theta=2 n \pi \pm \frac{\pi}{3}, n \in I$