Strengthening solve



Oops...to_poly returns the restriction sin(x) # 0, but this
condition is either ignored or mishandled:

(%i4) load("to_poly_solver")$

(%i13) to_poly_solve((x-%pi)/sin(x), x);
1 Enter to_poly[(x-%pi)/sin(x),[x]]
1 Exit  to_poly[[x-%pi],[sin(x)#0],[]]

(%o13) %union([x=%pi])

I thought to_poly_solve checked for such things.

Humiliated,
Barton

maxima-bounces at math.utexas.edu wrote on 09/01/2009 12:53:57 PM:

> looking at making Maxima's solve program more robust, maybe along the
> lines of Mathematica's Reduce, I came up with this simple problem,
> solve((x-%pi)/sin(x)=0, x)