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

$2{\sin }^2\theta -\cos 2\theta =0\Rightarrow 1-\cos 2\theta =0\Rightarrow \cos 2\theta =\frac{1}{2}$
$2\theta =\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$
$2{\cos }^2\theta -3\sin \theta =0$
$(2\sin \theta -1)(\sin \theta +2)=0\Rightarrow \sin \theta =\frac{1}{2}$
$\theta =\frac{\pi }{6},\frac{5\pi }{6}\Rightarrow No.ofsolutions=2$