Question

Solve the equation on the interval $0\le \theta \le 2\mathrm{\pi }$.

role="math" localid="1646734122417" $\sin \theta +\sin \left(2\theta \right)=0$

Short answer

On solving the equation, we get,

$\left\{0,\frac{2\mathrm{\pi }}{3},\mathrm{\pi },\frac{4\mathrm{\pi }}{3}\right\}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} \sin \theta +\sin \left(2\theta \right)=0 \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

We know, $\sin \left(2\theta \right)=2\sin \theta ·\cos \theta$,

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

Then,

$\sin \theta =0$

Also,

$$\begin{gathered} 1+2\cos \theta =0 \\ \cos \theta =-\frac{1}{2} \end{gathered}$$

Step 2. Solve the equation for the given interval.

Consider the given question,

$$\begin{gathered} \sin \theta =0 \\ \cos \theta =-\frac{1}{2} \\ 0\le \theta \le 2\mathrm{\pi } \end{gathered}$$

Solving $\sin \theta$for the interval $0\le \theta \le 2\mathrm{\pi }$,

$$\begin{gathered} \sin \theta =0 \\ \theta ={\sin }^{-1}\left(0\right) \\ \theta =0,\mathrm{\pi } \end{gathered}$$

Solving $\cos \theta =-\frac{1}{2}$for the interval $0\le \theta \le 2\mathrm{\pi }$,

$$\begin{gathered} \cos \theta =-\frac{1}{2} \\ \theta ={\cos }^{-1}\left(-\frac{1}{2}\right) \\ \theta =\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3} \end{gathered}$$

Therefore, the solution set is $\theta \in \left\{0,\frac{2\mathrm{\pi }}{3},\mathrm{\pi },\frac{4\mathrm{\pi }}{3}\right\}$.