Question

The number of solutions of the pair of equations $2{\sin }^2\theta -\cos 2\theta =0$ $2{\cos }^2\theta -3\sin \theta =0$ in the interval $[0,2\pi ]$ is

• A. 1
• B. 2
• C. 4

Answer 1

The correct option is B 2

$2{\sin }^2\theta -\cos 2\theta =0$
or $1-2\cos 2\theta =0$
or $\cos 2\theta =\frac{1}{2}$
$\Rightarrow 2\theta =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}$
or $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
where $\theta \in [0,2\pi ].$
Also $2{\cos }^2\theta -3\sin \theta =0$
or $2{\sin }^2\theta +3\sin \theta -2=0$
or $(2\sin \theta -1)(\sin \theta +2)=0$
or $\sin \theta =\frac{1}{2}[\because \sin \theta \ne 2]$
$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\text{where}\theta \in [0,2\pi ]$
combining Eqs. (i) and (ii), we get $\theta =\frac{\pi }{6},\frac{5\pi }{6}$
Therefore, there are two solutions.