Assume a population in Hardy-Weinberg equilibrium for a character trait with these genotypic frequencies: AA =?

There wouldn't be any. Make a Punnett square to see why. If you're removing the homozygous dominants (AA) you're only left with Aa and aa as options.

___a_|__a__
A| Aa |  Aa
a| aa  | aa

You can't get AA when one parent is homozygous recessive.

Try this:

Say there is a population of 100 with those frequencies:  25, 50, 25.  We'll remove the dominant homozygotes (25) leaving a population of 75.  The frequency of aa is now 0.33 and of Aa is 0.67.  So, we can do Hardy-Weinberg calculations based on a frequency of 0.33 of aa, which means the frequency of a is 0.57, which we let = q;  the frequency of A is 0.43 = p.

Hardy-Weinberg equation #2 says that the frequency of AA is now 0.18, the frequency of Aa is now 0.49, and teh frequency of aa is 0.32.

The frequency of dominant homozygotes is 0.18;  dominant homozygosity isn't as popular as it used to be...