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?
Best Answer
Each non-zero polar coordinate $(r,\theta)$ has an equivalent form $(-r,\theta+\pi)$. Applying this substitution to the two equations yields the new equations $r=-\cos2\theta$ and $r=\cos\theta+1$. Writing $\cos\theta=s$, these become $$r=2s^2-1\tag1$$ $$r=1-2s^2\tag2$$ $$r=s-1\tag3$$ $$r=s+1\tag4$$ The four combinations $(\{1,2\},\{3,4\})$ of these equations need to be considered separately, and yield the following valid solutions $(r,s)$ where $s\in[-1,1]$: $$(1,3)\to (-1,0),\left(-\frac12,\frac12\right)$$ $$(1,4)\to\left(\frac{5-\sqrt{17}}4,\frac{1-\sqrt{17}}4\right)$$ $$(2,3)\to\left(\frac{\sqrt{17}-5}4,\frac{\sqrt{17}-1}4\right)$$ $$(2,4)\to(1,0),\left(\frac12,-\frac12\right)$$ Negating both numbers in a pair above yields the same point by the aformentioned equivalence of polar coordinates (for $\cos(\pi+\theta)=-\cos\theta$), so we have the following distinct pairs with $r>0$: $$(1,0),\left(\frac12,-\frac12\right),\left(\frac{5-\sqrt{17}}4,\frac{1-\sqrt{17}}4\right)$$ The first two pairs above correspond to the four non-origin solutions already found: $(r,\theta)=\left(1,\pm\frac\pi2\right),\left(\frac12,\pm\frac{2\pi}3\right)$. The last one corresponds to the missing two solutions $(r,\theta)=\left(\frac{5-\sqrt{17}}4,\pm\cos^{-1}\frac{1-\sqrt{17}}4\right)$. The last intersection of the two polar curves is the origin.