#### Answer

The two points of intersection are: $(1, \frac{\pi}{4})$ and $(-1, \frac{3\pi}{4})$

#### Work Step by Step

$r = sin~2\theta$
$r = \sqrt{2}~cos~\theta$
To find the points of intersection, we can equate the expressions for $r$:
$sin~2\theta = \sqrt{2}~cos~\theta$
$2~sin~\theta~cos~\theta = \sqrt{2}~cos~\theta$
$2~sin~\theta = \sqrt{2}$
$sin~\theta = \frac{\sqrt{2}}{2}$
$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$
We can find $r$ when $\theta = \frac{\pi}{4}$:
$r = sin~2\theta$
$r = sin~\frac{2\pi}{4}$
$r = sin~\frac{\pi}{2}$
$r = 1$
We can find $r$ when $\theta = \frac{3\pi}{4}$:
$r = sin~2\theta$
$r = sin~\frac{6\pi}{4}$
$r = sin~\frac{3\pi}{2}$
$r = -1$
The two points of intersection are: $(1, \frac{\pi}{4})$ and $(-1, \frac{3\pi}{4})$