Here is an expanded take on my comment:

Imagine that we have a triangle of the sort you seek and the base angles are $\beta=\frac{p}{q}2\pi$ with the fraction in lowest terms (and $q \ge 6$). Then $x=(2cos\beta)a$

We know that $2a \cos \beta$ is an integer for $q=6.$ Otherwise it is an algebraic integer of degree $\frac{\phi(q)}{2}$ (short proof below).

So a not totally satisfactory answer to

Given a value $x$, determine if it can be a base and, if so, find all isosceles triangles over that base such that
side length $a$ is an integer

is: If $x$ is rational then it must be an integer and it arises only from an equilateral triangle. Otherwise,

- Verify that $x$ is an algebraic integer and determine the degree $d.$ (This means that you somehow know an exact value of $x.$)
- Find the (finite list) of integers with Euler totient function $\phi(q)=2d.$ (See the comment on this below.)
- Compute $2 cos(\frac{2p \pi}{q})$ for all cases.
- Check if $x$ is an integral multiple of any of them (and if so, of which). Voila!

That does not immediately shed any light (that I can see) on questions such as "Is there an $x$ which arises in over $7$ ways." But it does suggest picking a value $2d$ so that there are numerous solutions to $\phi(q)=2d$ and then computing the various $cos(\frac{2p \pi}{q})$ and looking for rational ratios. Maybe some nice patterns will become clear. (What solutions do you have, if one may ask.)

The same procedure offers, in some sense, an answer to the modified question

For which $x$, *algebraic of degree* $d$, are there solutions?

If one wanted a huge $d$ that might not be very practical.

Further comments:

Maple has the function invphi to compute the solutions of $\phi(q)=2d$. It wasn't always perfect, but that particular bug has been fixed. See also this question .

The fact about the degree of $2 \cos \beta$ must be a standard fact involving Chebyshev polynomials, but this seems to work:
Let $\zeta=e^{\frac {2\pi i}{q}}.$ Then $2\cos\beta=\zeta^p+\zeta^{-p}$ is a sum of two algebraic integers and hence itself an algebraic integer. The dimension of $\mathbb{Z}[\zeta^p]=\mathbb{Z}[\zeta]$ relative to $\mathbb{Z}$ is $\phi(q)$ but $\mathbb{Z}[\zeta^p]$ has dimension $2$ over $\mathbb{Z}[2\cos\beta] \subset \mathbb{R}$ as $t=\zeta^p$ satisfies $t^2-2\cos\beta t+1=0.$

I am not sure that you can have an $x$ in two ways in the isosceles case and feel that it might be fairly straightforward . But I have no proof. Do you have any examples? There seem to be none of degrees $4$ or $6$.

Yes, I am cascading the structure, building upon itself. – Joseph O'Rourke Jun 25 '13 at 16:24