The question was asking for the intersection points of $r=1+\cos \theta$ and $r=1-\cos \theta$ with $0≤ \theta ≤2\pi$, but doing:
$1+\cos \theta=1-\cos\theta$
0=2cosθ
0=cosθ
θ=$\frac π2$ or $\frac{3\pi}{2}$
Yields an incomplete answer.
When the two curves are graphed a third intersection appears at the pole, but at two separate angles for each graph.
θ=π and θ=0
How would I find this intersection point without looking at the graph and tracing the curves?
Best Answer
You need not just solve for $f(\theta) = g(\theta)$, but also solve for $f(\theta) = -g(\pi + \theta)$, to take into account the chance that a point of intersection of this form.
For your example $f = 1+\cos\theta, g = 1-\cos\theta$, so you need to solve for $1+\cos\theta = -1+\cos(\theta+\pi) = -1-\cos\theta$. This simplifies to $\cos\theta = -1$, so $\theta = \pi$, as desired.
edit: I just realized that you generally need to check the origin as a special case, too, since $r,\theta$ represents the origin whenever $r=0$ no matter the value of $\theta$.