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

The correct option is C two

$2{\sin }^2\Theta -\cos 2\Theta =0\Rightarrow 2{\sin }^2\Theta -(1-2{\sin }^2\Theta )=0\Rightarrow {\sin }^2\Theta =\frac{1}{4}\Rightarrow \sin \Theta =\pm \frac{1}{2}$
also $2{\cos }^2\Theta =3\sin \Theta \Rightarrow 2{\sin }^2\Theta +3\sin \Theta -2=0\Rightarrow \sin \Theta =\frac{1}{2}$
Hence common solution is, $\sin \Theta =\frac{1}{2}\Rightarrow \Theta =\frac{\pi }{6},\frac{5\pi }{6}\Rightarrow$ two solutions in $[0,2\pi ]$.