Reduce[{4==Abs[-1+2 Cos[2A]+2 Cos[2B]+2 Cos[2(A+B)]]},{A,B},Reals]
worked well, but
Reduce[{4==Abs[-1+2 Cos[2A]+2 Cos[2B]+2 Cos[2(A+B)]],0<A<Pi,0<B<Pi,0<A+B<Pi},{A,B},Reals]
given Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>
Reduce[{4 == Abs[-1 + 2 Cos[2 A] + 2 Cos[2 B] + 2 Cos[2 (A + B)]], 0 < B < Pi}, {A, B}, Reals]raisesReduce::nsmet. Replace0 < B < Piwith0 < A < Pidoes not. The expression is symmetric inAandB. – Patrick Stevens Aug 19 '15 at 11:36Reduce[{4 == Abs[-1 + 2 Cos[2 A] + 2 Cos[2 B] + 2 Cos[2 (A + B)]], 0 < B < Pi}, {A, B}, Reals]returns a solution; the seemingly equivalentReduce[{4 == Abs[-1 + 2 Cos[2 A] + 2 Cos[2 B] + 2 Cos[2 (A + B)]], 0 < B < Pi}, {B, A}, Reals]returnsReduce::nsmet! (on MMA 10.2 Win7-64) – MarcoB Aug 19 '15 at 16:26