Question

1. 0.25/1 points details my notes previous answers ask your teacher practice another scalcet9 10.4.039 find all points of intersection of the given curves. (assume ( 0 leq \theta leq 2pi ) and ( r geq 0 ). order your answers from smallest to largest ( \theta ). if an intersection occurs at the pole, enter pole in the first answer blank.)( r = 8sin(2\theta) ), ( r = 4 )( (r, \theta) = left(4, \frac{pi}{12}

ight) )

Explanation:

Step1: Set the equations equal

Set \(8\sin(2\theta)=4\).

$$\begin{align*} 8\sin(2\theta)&=4\\ \sin(2\theta)&=\frac{4}{8}=\frac{1}{2} \end{align*}$$

Step2: Solve for \(2\theta\)

We know that if \(\sin x = a\), then \(x = n\pi+(- 1)^{n}\arcsin(a)\), \(n\in\mathbb{Z}\). For \(\sin(2\theta)=\frac{1}{2}\), \(2\theta = n\pi+(-1)^{n}\frac{\pi}{6}\).
Since \(0\leq\theta\leq2\pi\), then \(0\leq2\theta\leq4\pi\).
When \(n = 0\), \(2\theta=\frac{\pi}{6}\), so \(\theta=\frac{\pi}{12}\); when \(n = 1\), \(2\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\), so \(\theta=\frac{5\pi}{12}\); when \(n = 2\), \(2\theta = 2\pi+\frac{\pi}{6}=\frac{13\pi}{6}\), so \(\theta=\frac{13\pi}{12}\); when \(n = 3\), \(2\theta=3\pi - \frac{\pi}{6}=\frac{17\pi}{6}\), so \(\theta=\frac{17\pi}{12}\)

Answer:

\(\theta=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}\)