Chapter H.2, Problem 29E

To determine

To find: The all the points of the given curves.

Answer to Problem 29E

  $\left(1,\frac{\pi }{12}\right),\left(1,\frac{11\pi }{12}\right),\left(-1,-\frac{\pi }{12}\right),\left(-1,-\frac{11\pi }{12}\right),\left(1,\frac{5\pi }{12}\right),\left(1,\frac{7\pi }{12}\right),\left(-1,-\frac{5\pi }{12}\right),\left(-1,-\frac{7\pi }{12}\right)$

Explanation of Solution

Given: The curves are $r=2\sin 2\theta \text{ and }r=1$ .

First draw the both curves then find the intersection point.

There are total 8 intersections points.

  $\begin{array}{c}2\sin 2\theta =1\end{array}$

  $\begin{array}{c}\sin 2\theta =\frac{1}{2}\end{array}$

  $\begin{array}{c}\sin 2\theta =\sin \frac{\pi }{6}\end{array}$

  $\begin{array}{c}2\theta =2n\pi \pm \frac{\pi }{6}\end{array}$

  $\begin{array}{c}\theta =n\pi \pm \frac{\pi }{12}\end{array}$

The intersections points are

  $\left(1,\frac{\pi }{12}\right),\left(1,\frac{11\pi }{12}\right),\left(-1,-\frac{\pi }{12}\right),\left(-1,-\frac{11\pi }{12}\right),\left(1,\frac{5\pi }{12}\right),\left(1,\frac{7\pi }{12}\right),\left(-1,-\frac{5\pi }{12}\right),\left(-1,-\frac{7\pi }{12}\right)$