I just recently asked about limacon curves, but now I want to examine the Archimedian curve:
$$r=a + b\theta$$
The question is essentially the same, for a given range, how would one find the minimum $x$-value in a particular revolution. Take $r=1 - \frac{0.5}{2\pi}\theta$ for $0\le\theta<360$:
Ignore the fact that the picture goes to $3\pi$.
It is very slight and hard to tell, but our smallest $x$ value is NOT at $\theta=180$. But where is it? Obviously it is less that $180$, but how do I get that angle/point? I have already tried differentiating my function, but it does not work out to be a nice solve.
Here's the Lagrange multiplier approach:
We are trying to minimize $f(r,\theta) = r \cos \theta$ subject to the constraint $g(r,\theta) = r - b\,\theta = a$.
We calculate $$ \nabla f = (\cos \theta,-r\sin \theta)\\ \nabla g = (1,-b) $$ So, setting $\nabla f = \lambda \nabla g$ and $g(r,\theta) = a$, we have the system of equations $$ \cos \theta = \lambda\\ -r \sin \theta = -b \lambda\\ r - b \theta = a $$ Any $(r,\theta)$ solving the above system will be a local minimum or a local maximum for the $x$-coordinate.
Attempted solution: substituting $\lambda = \cos \theta$, the second equation yields $$ -r \sin \theta = -b \cos \theta \implies r = b\cot \theta $$ From there, we may make the final equation an equation on $\theta$ alone: $$ b \cot \theta - b\theta = a \implies\\ \cot \theta - \theta = \frac ab $$ This equation cannot be solved directly with any commonly used algebraic methods, but solutions can easily be found via numerical methods.
NOTE: I seem to have chosen the wrong critical point below. I will correct it if I can. Ignore for now.
For the example you've provided, $a = 1$, and $b = -1/(4 \pi)$. According to WA, the third positive critical value of $\theta$ (i.e. the one you want) is $$ \theta \approx 6.12907243518723 \approx (0.975)(2 \pi) $$