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. $\theta =2n\pi \pm ,\frac{\pi }{3}n\in I$
• C. $\theta =2n\pi \pm \frac{\pi }{6},n\in I$
• 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
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3},n\in I$