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Mar 6 at 4:05 history edited Bob Hanlon CC BY-SA 4.0
Added purely graphic solution
Mar 5 at 20:50 comment added user64494 Solve[{eqn, -2 < x < 0}, x,Reals] works in the end, taking a dozen of minutes.
Mar 5 at 20:44 comment added Bob Hanlon Use Solve[{eqn, -2 < x < 0}, x, Complexes] As shown in the documentation, "Solve[expr && vars [Element] Reals, vars, Complexes] solves for real values of variables, but function values are allowed to be complex." In this specific case a tighter constraint than vars \[Element] Reals is used.
Mar 5 at 20:26 comment added user64494 -1. I'd like to add that Solve[{eqn, -2 < x < 0}, x] produces "Solve::incs: Warning: Solve was unable to prove that the solution set found is complete. {{x -> -(9/5)}, {x -> -1}}". Your NegativeReals stands to disguise it.
Mar 5 at 20:04 comment added user64494 In fact , you find FunctionDomain[2 ArcCos[Sin[Pi *x]] - Pi *Cos[ArcSin[x + 1]] over the reals (which is exactly x>=-2&&x<=0), making use of a plot ("From the plot, restrict the domain"). This is my answer in other formulas. Because of this reason I can neither up vote nor accept it.
Mar 5 at 19:53 history edited Bob Hanlon CC BY-SA 4.0
Added FindInstance
Mar 5 at 19:47 history answered Bob Hanlon CC BY-SA 4.0