Computing planet conjunctions with 2D circular orbits still hard?

All methods I’ve seen for computing planetary/stellar conjuctions
(when two planets or a planet and a star appear close together in the
sky because their angular separation, as measured from Earth, is
small) are iterative: they look at the positions day by day and find
the smallest angular separation.

EDIT (per comment): All of the constants (a1, b1, a2, b2, a3, c3) are known values. I’m trying to solve for t given these constants. There are, of course, an infinite number of solutions, so solutions for t in any given interval are fine too. Numerical solutions (where I plugin the constants as approximate numbers) are fine too, provided they are reasonably efficient (moreso than iterating).

I thought it would be easier with circular 2-dimensional orbits, so I
tried this:

 
o[t_] = {Cos[t],Sin[t]} 
o1[t_] = a1*{Cos[a2*t+a3],Sin[a2*t+a3]} 
o2[t_] = b1*{Cos[b2*t+b3],Sin[b2*t+b3]} 

The above are the circular 2-dimensional orbit of 3 planets, where one
planet’s orbit has been normalized to have radius 1, period 2*Pi, and
o[0] = {1,0}.

Treating “o” as Earth, the position of “o1” as viewed from Earth at
time t is “o1[t]-o[t]”, and “o2[t]-o[t]” for “o2”.

The angles from Earth are “Apply[ArcTan,o1[t]-o[t]]” and
“Apply[ArcTan,o2[t]-o[t]]”. The difference between these angles is:

 
Abs[Apply[ArcTan,o1[t]-o[t]]-Apply[ArcTan,o2[t]-o[t]]] 

which is what I’m trying to minimize. In fact, because we’re in 2
dimensions, the minimum value will be 0, so I just need to find the
zeros of the function above (of course, the Abs[] becomes superfluous
at this point).

I’ve tried several approaches, and nothing worked. Solve[] won’t solve
the equation for Reals and hangs when I don’t give it a domain.

I’ve tried using dot products, and even tried solving the simpler
problem of when two planets appear perpendicular (dot product is 0),
to no avail.

Graphing with actual vales show there’s no real pattern to the zeros
of my function, or even a minimum distance between zeros.

I could try an iterative solution of some sort, but then I’m back to
square one.