Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

Given pair of equations $2{\sin }^2\theta -\cos 2\theta =0$ and $2{\cos }^2\theta -3\sin \theta =0$
$2{\sin }^2\theta -\cos 2\theta =0$
$\Rightarrow 2{\sin }^2\theta -(1-2{\sin }^2\theta )=0$
$\Rightarrow 4{\sin }^2\theta =1$
$\Rightarrow \sin \theta =\pm \frac{1}{2}$
And $2{\cos }^2\theta -3\sin \theta =0$
$\Rightarrow 2(1-{\sin }^2\theta )-3\sin \theta =0$
$\Rightarrow 2{\sin }^2\theta +3\sin \theta -2=0$
$\Rightarrow \sin \theta =\frac{-3\pm \sqrt{9+16}}{4}$
$\Rightarrow \sin \theta =\frac{-3\pm 5}{4}$
$\Rightarrow \sin \theta =\frac{1}{2}as-1\le sin\theta \le 1$
So $\sin \theta =\frac{1}{2}$ satisfies both the equations
$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$ as $\theta \in [0,2\pi ]$