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

$2$

Find the number of solutions of the pair of equations

Given, $2{\sin }^2\theta -\cos 2\theta =0$and $2{\cos }^2\theta -3\sin \theta =0$

We have,

$$\begin{gathered} 2{\sin }^2\theta –\cos 2\theta =0 \\ \Rightarrow 2{\sin }^2\theta –(1–2{\sin }^2\theta )=0 \\ \Rightarrow 2{\sin }^2\theta –1+2{\sin }^2\theta =0 \\ \Rightarrow 4{\sin }^2\theta =1 \\ \Rightarrow {\sin }^2\theta =\frac{1}{4} \\ \Rightarrow \sin \theta =\frac{1}{2},\frac{-1}{2}...\left(\mathrm{i}\right) \end{gathered}$$

Now,

$$\begin{gathered} 2{\cos }^2\theta -3\sin \theta =0 \\ \Rightarrow 2(1–{\sin }^2\theta )–3\sin \theta =0 \\ \Rightarrow 2–2{\sin }^2\theta –3\sin \theta =0 \\ \Rightarrow 2{\sin }^2\theta +3\sin \theta –2=0 \\ \Rightarrow 2{\sin }^2\theta +4\sin \theta –\sin \theta –2=0 \\ \Rightarrow 2\sin \theta \left(\sin \theta +2\right)-1\left(\sin \theta +2\right)=0 \\ \Rightarrow \left(2\sin \theta -1\right)\left(\sin \theta +2\right)=0 \end{gathered}$$

$\Rightarrow \sin \theta =\frac{1}{2},\sin \theta =-2$ (not possible) $...\left(\mathrm{ii}\right)$

From (i) and (ii),

$\theta =\frac{\pi }{6},\frac{5\pi }{6}$

Therefore, the number of solutions $=2$

Hence, option (C) is the correct answer.