Question

The number of solutions of $\cos \left(2\theta \right)=\sin \left(\theta \right)$ in $\left(0,2\pi \right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C

$3$

Explanation for correct option

Given, $\cos \left(2\theta \right)=\sin \left(\theta \right)$

$$\begin{gathered} \Rightarrow 1-2{\sin }^2\left(\theta \right)=\sin \left(\theta \right)\left[\because \cos \left(2x\right)={\cos }^2\left(x\right)-{\sin }^2\left(x\right)\&{\cos }^2\left(x\right)+{\sin }^2\left(x\right)=1\right] \\ \Rightarrow 2{\sin }^2\left(\theta \right)+\sin \left(\theta \right)-1=0 \\ \Rightarrow 2{\sin }^2\left(\theta \right)+2\sin \left(\theta \right)-\sin \left(\theta \right)-1=0 \\ \Rightarrow 2\sin \left(\theta \right)\left(\sin \left(\theta \right)+1\right)-\left(\sin \left(\theta \right)+1\right)=0 \\ \Rightarrow \left(\sin \left(\theta \right)+1\right)\left(2\sin \left(\theta \right)-1\right)=0 \end{gathered}$$

$\therefore \left(\sin \left(\theta \right)+1\right)=0$ or $\left(2\sin \left(\theta \right)-1\right)=0$

$\therefore \sin \left(\theta \right)=-1$ or $\sin \left(\theta \right)=\frac{1}{2}$

$\because \theta \in \left(0,2\mathrm{\pi }\right),\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$

Therefore there are $3$ solutions of $\theta$

Hence, the correct option is $\left(\mathrm{C}\right)$.