how many bitangents on this hypotrochoid? After playing with spirograph, a bit I realized all these curves I'm drawing should be an algebraic curve and it's birational equivalent to a $\mathbb{P}^1$.  In the example below, I have a six-sided hypotrochoid.  The equation is $z = e^{i\theta} ( \frac{7}{8} + \frac{1}{4} e^{6i\theta}) $, so this already gives a (birational?) map from $S^1 \to \mathbb{C}$.
I learned one way to estimate the degree of this map is to intersect a line (or hyperplane divisor) and computing the number of intersection point (the degree)
Typical solutions in algebraic geometry involve deforming the problem to one that is more generic. Or complexifying the problem. 
The intersection I am looking for (a bitangent) is not generic and I believe this curve is singular (at least it has self-intersections). 
So far, I have counted about 48 or 50 (by literally) drawing with a pencil. I will try to sketch my partial results. This question certainly has a finite, closed-ended meaningful answer.  This certainly falls  under classical enumerative algebraic geometry (such as the Schubert calculus).

 A: Too long for a comment. Using GKZ p. 17, a parametric representation of the dual curve is: $$X={\frac {8}{7}}\,{\frac {\cos \left( \theta \right)  \left( -7+128\,
 \left( \cos \left( \theta \right)  \right) ^{6}-224\, \left( \cos
 \left( \theta \right)  \right) ^{4}+112\, \left( \cos \left( \theta
 \right)  \right) ^{2} \right) }{288\, \left( \cos \left( \theta
 \right)  \right) ^{2}+512\, \left( \cos \left( \theta \right) 
 \right) ^{6}-768\, \left( \cos \left( \theta \right)  \right) ^{4}-5}
}$$
$$Y=8\,{\frac { \left( -1+128\, \left( \cos \left( \theta \right) 
 \right) ^{6}-160\, \left( \cos \left( \theta \right)  \right) ^{4}+48
\, \left( \cos \left( \theta \right)  \right) ^{2} \right) \sin
 \left( \theta \right) }{288\, \left( \cos \left( \theta \right) 
 \right) ^{2}+512\, \left( \cos \left( \theta \right)  \right) ^{6}-
768\, \left( \cos \left( \theta \right)  \right) ^{4}-5}}$$
And as @MarianoSuárez-Álvarez says; nodes $(X,Y)$ corresponds to bitangents $Xx+Yy+1=0$
 to your curve.

A: I've never posted here before and meant to post this as a comment but accidentally posted it as an answer.  Apologies.
A very naive, elementary approach:  No line through the origin is a bitangent, so every bitangent has a leftmost point of tangency as viewed from the origin.  Fix an arc A such that the curve is the disjoint union of A and 5 congruent copies of A.  Count all bitangents whose leftmost point of tangency is on A and multiply by 6.
To count all bitangents whose leftmost point of tangency is on A, consider a point P which moves from one end of A to the other, and note all possible positions for P for which the tangent at P could also be a tangent somewhere to the right of P.
Robert Hill
