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. zero
• B. one
• C. two
• D. four

Answer 1

Answer: (C)

two

Given $2{\sin }^2\theta -\cos 2\theta =0$ and $2{\cos }^2\theta -3\sin \theta =0$
$\Rightarrow 2{\sin }^2\theta -(1-2{\sin }^2\theta )=0$ and $2(1-{\sin }^2\theta )-3\sin \theta =0$
$\Rightarrow 4{\sin }^2\theta -1=0$ and $2{\sin }^2\theta +3\sin \theta -2=0$
$\Rightarrow \sin \theta =\pm \frac{1}{2}$ and $\sin \theta =-2,\frac{1}{2}$
$\therefore \sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$ in the given interval.