Question

Solve :

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

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

Answer 1

The correct option is B $\theta =2n\pi +\frac{\pi }{3},n\in I$

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

we know that

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$\cos 2\theta =2{\cos }^2\theta -1$

$\Rightarrow \frac{4{\cos }^3\theta -3\cos \theta }{2(2{\cos }^2\theta -1)-1}=\frac{1}{2}$

$\Rightarrow \frac{\cos \theta (4{\cos }^2\theta -3)}{4{\cos }^2\theta -2-1}=\frac{1}{2}$

$\Rightarrow \frac{\cos \theta (4{\cos }^2\theta -3)}{(4{\cos }^2\theta -3)}=\frac{1}{2}$

$\Rightarrow \cos \theta =\frac{1}{2}$ ${\cos }^2\theta \ne \frac{3}{4}$

$\Rightarrow \cos \theta =\cos \frac{\pi }{3}$

We know that if $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha$

$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3}$