Question

In Problems 11–34, solve each equation on the interval $0\le \theta \le 2\pi$.

$2\sin \theta +3=2$

Short answer

The solution set is$\left\{\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}\right\}$.

Step-by-step solution

Step 1. Given Information

In the given problem we have to solve each equation on the interval$0\le \theta \le 2\pi$.

$2\sin \theta +3=2$

Step 2. Firstly solving the equation 2sinθ+3=2

Subtract by 3 on both side

$$\begin{gathered} 2\sin \theta +3-3=2-3 \\ 2\sin \theta =-1 \end{gathered}$$

Divide by 2 on both side

$$\begin{gathered} \frac{2}{2}\sin \theta =-\frac{1}{2} \\ \sin \theta =-\frac{1}{2} \end{gathered}$$

Step 3. After solving the equation sinθ=-12

In the interval $[0,2\mathrm{\pi })$, there are two angles $\theta$for which

localid="1646564805462" $\sin \theta =-\frac{1}{2}:\theta =\frac{7\mathrm{\pi }}{6}\mathrm{and}\theta =\frac{11\mathrm{\pi }}{6}$

So the solution set is $\left\{\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}\right\}$.