Question

The number of values of $\theta$ in the interval $[-\pi ,\pi ]$ satisfying the equations $\cos \left(\theta \right)+\sin \left(2\theta \right)=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for the correct answer:

We are given,

$$\begin{gathered} \cos \left(\theta \right)+\sin \left(2\theta \right)=0 \\ \Rightarrow \cos \left(\theta \right)+2\sin \left(\theta \right)\cos \left(\theta \right)=0 \\ \Rightarrow \cos \left(\theta \right)(1+2\sin \theta )=0 \\ \Rightarrow \cos \left(\theta \right)=0,1+2\sin \left(\theta \right)=0 \\ \Rightarrow \cos \left(\theta \right)=0,\sin \left(\theta \right)=\left(-\frac{1}{2}\right) \end{gathered}$$

Case 1: $\cos \left(\theta \right)=0$

$\cos \left(\theta \right)=\cos \left(\frac{\mathrm{\pi }}{2}\right)$

$\theta =-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}$ {since $\theta$ lies in $[-\pi ,\pi ]$}

Case 2: $\sin \left(\theta \right)=-\frac{1}{2}$

$$\begin{gathered} \sin \left(\theta \right)=\sin \left(-\frac{\mathrm{\pi }}{6}\right)=\sin \left(\frac{\mathrm{\pi }}{6}-\mathrm{\pi }\right) \\ \Rightarrow \theta =-\frac{\mathrm{\pi }}{6},-\frac{5\mathrm{\pi }}{6} \end{gathered}$$

There are four solutions.

Hence, option D is the correct answer.