Question

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

$\sin \left(3\theta \right)=-1$

Short answer

The solution set is$\left\{\frac{\mathrm{\pi }}{2},\frac{7\mathrm{\pi }}{3},\frac{11\mathrm{\pi }}{3}\right\}$.

Step-by-step solution

Step 1. Given Information 

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

$\sin \left(3\theta \right)=-1$

Step 2. In the interval [0,2π), the sine function equals -1 at 3π2.

So, we know that $3\theta$must equal $\frac{3\mathrm{\pi }}{2}$.

To find these solutions, write the general formula that gives all the solutions.

$3\theta =\frac{3\mathrm{\pi }}{2}+2\mathrm{\pi n}$

Divide by 3 on both side

localid="1646585218967" $$\begin{gathered} \frac{3}{3}\theta =\frac{1}{3}\left(\frac{3\mathrm{\pi }}{2}+2\mathrm{\pi n}\right) \\ \theta =\frac{1}{3}·\frac{3\mathrm{\pi }}{2}+2\mathrm{\pi n}·\frac{1}{3} \\ \theta =\frac{\mathrm{\pi }}{2}+\frac{2\mathrm{\pi n}}{3} \end{gathered}$$

Step 3. The general formula is θ=π2+2πn3

So the value of given function in interval $[0,2\mathrm{\pi })$is

$\theta =\frac{\mathrm{\pi }}{2}+\frac{2\mathrm{\pi n}}{3}$

localid="1646586125845" role="math" $$\begin{gathered} \theta =\frac{\mathrm{\pi }}{2}+\frac{2\pi \times 0}{3}\theta =\frac{\mathrm{\pi }}{2}+\frac{2\pi \times 1}{3}\theta =\frac{\mathrm{\pi }}{2}+\frac{2\pi \times 2}{3} \\ \theta =\frac{\mathrm{\pi }}{2}+\frac{0}{3}\theta =\frac{3\mathrm{\pi }+4\pi }{3}\theta =\frac{3\mathrm{\pi }+8\pi }{3} \\ \theta =\frac{\mathrm{\pi }}{2}\theta =\frac{7\mathrm{\pi }}{3}\theta =\frac{11\mathrm{\pi }}{3} \end{gathered}$$

So the solution set islocalid="1646586141392" $\left\{\frac{\mathrm{\pi }}{2},\frac{7\mathrm{\pi }}{3},\frac{11\mathrm{\pi }}{3}\right\}$