The equations I wanted to solve is as follows
neweqns=Sin[a - b] == Sin[b] + Sin[b - d] && Sin[c] + Sin[c - d] == 0 &&
Sin[a - d] + Sin[b - d] + Sin[c - d] == Sin[d] &&
Sin[a] + Sin[b] + Sin[c] + Sin[d] ==
0 && ((Sin[d] == 0 && 2 Sin[b] + Sin[c] == 0 &&
2 Sin[a] + Sin[c] == 0) || (Sin[d] == 0 && Sin[c] == 0 &&
Sin[a] + Sin[b] ==
0) || ((Sqrt[3] + 2 Sin[d] == 0 || 2 Sin[d] == Sqrt[3]) &&
Sin[c] + Sin[d] ==
0 && (Sqrt[3] + 2 Sin[b] == 0 || 2 Sin[b] == Sqrt[3]) &&
Sin[a] + Sin[b] ==
0) || ((Sqrt[3] + 2 Sin[d] == 0 || 2 Sin[d] == Sqrt[3]) &&
Sin[c] == Sin[d] && Sin[b] + Sin[d] == 0 &&
Sin[a] + Sin[d] ==
0) || ((Sin[d] == -((3 I Sqrt[5])/2) ||
Sin[d] == (3 I Sqrt[5])/2) && 3 Sin[c] + Sin[d] == 0 &&
3 Sin[b] + Sin[d] == 0 && 3 Sin[a] + Sin[d] == 0))
Reduce[neweqns && 0 <= a < 2 \[Pi] && 0 <= b < 2 \[Pi] &&
0 <= c < 2 Pi && 0 <= d < 2 Pi, {a, b, c, d},
Reals] // FullSimplify
However, I got the following error:
I have three questions about this:
- How to avoid such error and obtain the solutions?
- Should I add constraints on constants evolved in the solving process (add terms, eliminate terms, multiply terms, etc.) to be Reals? Will this active delete something which actually are solutions?
- What does
<<1>>
mean in the error information?
Thanks a lot for any suggestion!
Reals
. Just remove the domain restriction. You could alternatively usesol = Solve[neweqns && 0 <= a < 2 π && 0 <= b < 2 π && 0 <= c < 2 π &&0 <= d < 2 π, {a, b, c, d}, Method -> Reduce] // FullSimplify
$\endgroup$Reals
, it works. I am still confusing the reason.. Does domain mean the domain of all variables? $\endgroup$Reduce
, "If dom is Reals, or a subset such as Integers or Rationals, then all constants and function values are also restricted to be real." The presence ofI
causes a contradiction which is what the error message indicated. $\endgroup$