Deceptive intersection of polar curves Consider the graphs of $r=\cos 2\theta$ and $r=\cos\theta -1$.
I would like to find all the intersections for the two graphs.
Equating both and solve I only manage to find $(-1,\frac{\pi}{2}),(-1,\frac{3\pi}{2}),(-\frac{1}{2},\frac{\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3})$.
However, there seem to be other intersections as well as the two graphs may not meet at the same value of $\theta$.
Is there a way to solve for the other points of intersection?

$\cos 2\theta = \cos\theta - 1\\
2\cos^2 \theta - 1 = \cos\theta - 1\\
\cos\theta(2\cos\theta - 1) = 0\\
\theta = \frac {\pi}{2},\frac {3\pi}{2}, \frac {\pi}{3}, \frac {5\pi}{3}$
However, the curves will also overlap when $r(\theta)$ in one curve equals $-r(\theta + \pi)$ in the other.
$\cos 2\theta = \cos\theta + 1\\
2\cos^2 \theta - 1 = \cos\theta + 1\\
2\cos^2 \theta - \cos\theta - 2$
$\cos \theta = \frac {1 - \sqrt {17}}{4}$
Which will give two more points.
The curves $r = \cos \theta - 1$ and $r=\cos\theta + 1$ are identical curves.
And the curves intersect at $r = 0$
