@Carl Love What does `evala/toprof`(Pf); `evala/toprof`(indets(Pf,RootOf)); return? ANS2 should contain variable k. If it does not then what would explain this? Also I added a print command.

I am editing too much. The codes work flawlessly in Maple V R4. I wonder in which version the discontinuity occurs.

@Markiyan Hirnyk It appears _EnvExplicit:=true; is being used, converting all RootOfs into radicals. Lets see if _EnvExplicit:=false; would change that.

@Carl Love roots or evala@Roots or another? Which is better?

@Carl Love Thank you for your help. It appears roots command does not succeed. I'll change it to evala@Roots.

I ask, what command would you use to obtain roots of a factored polynomial?

@Carl Love I suppose split is the same thing as evala@AFactor. I tested it. It seems to work. I already edited my original answer.

@Dave L If you do that you should state the author of the statement you edit.

@Carl Love I have forgotten about this option completely, my apology. Unfortunately I have no access to the Groebner package. I can only suggest using method=buchberger and plex(X,Q). Quote from the help file for GB:

- GB is an implementation of Buchberger's algorithm for computing Grobner bases
  over Zp(a,b,...)[x1, ..., xn].
      
- GB(F,X,termorder) computes the reduced, minimal Grobner basis of the
  polynomials F with respect to the indeterminates X and the given term order-
  ing.      

No help otherwise.

..Also the basis you quoted is identical to tdeg basis.

@testht06 

Your definitions for x and q are:

so the monic for Q is what you seek and it is obtained with:

@roman_pearce Thank You, that is precisely the function I needed. But also I was wondering if there is any support for this from side of Maple (even though the solution is so simple).

Perhaps you want to try it using symmetric polynomials:

Look for symmetric polynomials package.

@Carl Love I was supposing this could work:

But now I think it would work only for small cases.

@Carl Love Maple 16 seems nicer.

What would this produce, 25 answers?

Also

gives full solution. So a solution would be to first figure each variable's range. In observation it seems a false approach for isolve to produce over a cartesian product of ranges.

Does isolve accept inert Product expressions?

Dividing (c,d) plane with the polynomials involved this image is obtained. The range of x: -1..1 was not used.

@Carl Love The order of eliminations I employed:

{cos(p1), sin(p1), sin(p2), cos(p3), sin(p3), cos(p2)}, {cos(t3), sin(t3)}, {sin(t2)}, {cos(t2)}, {y}.

Eliminating cos(t2) gave the first occurence of RootOf. Eliminating y gave 3 cases, all quadratic in x. That is my previous post. Perhaps you could reproduce this.