Question

The number of solutions of equation $6\cos 2\theta +2{\cos }^2\theta /2+2{\sin }^2\theta =0,-\pi <\theta <\pi$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is B $4$

Given, $6\cos 2\theta +2{\cos }^2\frac{\theta }{2}+2{\sin }^2\theta =0$
$\Rightarrow 6(2{\cos }^2\theta -1)+(1+\cos \theta )+(2-2{\cos }^2\theta )=0$
$\Rightarrow 10{\cos }^2\theta +\cos \theta -3=0$
$\Rightarrow (2\cos \theta -1)(5\cos \theta +3)=0$
$\Rightarrow \cos \theta =\frac{1}{2},\frac{-3}{5}$
Therefore solutions in the given interval are,
$\theta =-{\cos }^{-1}\left(\frac{-3}{5}\right),-\frac{\pi }{3},\frac{\pi }{3},{\cos }^{-1}\left(\frac{-3}{5}\right)$