5
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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. >>

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    $\begingroup$ I think this should have the [bugs] tag. Reduce[{4 == Abs[-1 + 2 Cos[2 A] + 2 Cos[2 B] + 2 Cos[2 (A + B)]], 0 < B < Pi}, {A, B}, Reals] raises Reduce::nsmet. Replace 0 < B < Pi with 0 < A < Pi does not. The expression is symmetric in A and B. $\endgroup$ – Patrick Stevens Aug 19 '15 at 11:36
  • $\begingroup$ Yes, definitely smells like a bug. Even more startling is the fact that the result depends on the order in which you specify the variables to solve for: Reduce[{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 equivalent Reduce[{4 == Abs[-1 + 2 Cos[2 A] + 2 Cos[2 B] + 2 Cos[2 (A + B)]], 0 < B < Pi}, {B, A}, Reals] returns Reduce::nsmet! (on MMA 10.2 Win7-64) $\endgroup$ – MarcoB Aug 19 '15 at 16:26
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    $\begingroup$ This may be related to Why does simplification in Mathematica depend on variable names and Why does Simplify ignore an assumption?. $\endgroup$ – MarcoB Aug 19 '15 at 16:28

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