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3

It seems that Mathematica can handle this when you switch to Exp form of expressions: ArcCos[3/5] + 2 ArcSec[Sqrt[5]] // TrigToExp // FullSimplify ArcCos[3/5] - 2 ArcCos[2/Sqrt[5]] // TrigToExp // FullSimplify π 0


1

I don't think this will work in every possible case but for your examples you could write a function doing : expr1 = ArcCos[3/5] + 2 ArcSec[Sqrt[5]]; expr2 = ArcCos[3/5] - 2 ArcCos[2/Sqrt[5]]; Solve[{Sin[x] == (Sin[expr1] // TrigExpand), Cos[x] == (Cos[expr1] // TrigExpand), -π < x <= π}, x, Reals] (* {{x -> π}} *) Solve[{Sin[x] == ...


5

Another way is to use Maximize rather than solving for the zero of the derivative. f[x_] = (r ω^2 Sin[x] Cos[x])/(g - ω^2 (Cos[x])^2 r); as = {r > 0, ω > 0, g > 0, r ω^2 < g}; You can see that f[x] // TrigReduce (* -((r ω^2 Sin[2 x])/(-2 g + r ω^2 + r ω^2 Cos[2 x])) *) therefore you can make a simple substitution and, assuming the ...



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