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3

It works if you exclude the singular point: Assuming[2 Pi > θ > 0, FullSimplify[1/Sqrt[1 + Abs[Cot[θ/2]]^2]]] (* Sin[θ/2] *)


1

The command InverseFunction[a Tanh[d (# - g)] + a/2 (Tanh[d (m + g)] - Tanh[d (m - g)]) &][y] gives the output $$ \frac{\tanh ^{-1}\left(\frac{a \tanh (d (m-g))-a \tanh (d (g+m))+2 y}{2 a}\right)+d g}{d}$$ together with the warning that $\tanh^{-1}$ is a multivalued function.


0

Basically the thing i realized is that, as it was pointed out by Timothy: ArcTan[x+I y] != ArcTan[x, y] == Arg[x+I y] You can try it out with numerical values and you'll see 1!=2==3. N[1/(2 \[Pi]) ArcTan[1/2 + I Sqrt[3]/2]] N[1/(2 \[Pi]) ArcTan[1/2, Sqrt[3]/2]] N[1/(2 \[Pi]) Arg[1/2 + I Sqrt[3]/2]] and you get 0.125 + 0.1048 I 0.166667 ...


3

NSolve (or Reduce as pointed out by @ciao in the comments above) will find all solutions in a specified interval $Version "10.2.0 for Mac OS X x86 (64-bit) (July 7, 2015)" NSolve[{5.20427*Sin[0.529699*t] == Sin[2*Pi*t], 0 <= t <= 100}, t] {{t -> 0.}, {t -> 0.32491}, {t -> 5.97889}, {t -> 11.6173}, {t -> 12.1353}, {t -> 12.2205}, {t ...


1

The package RootSearch, by Ted Ersek, does what you need : RootSearch[5.20427*Sin[0.529699*t] == Sin[2*Pi*t], {t, 0, 20}] { {t->0.}, {t->0.3249099866}, {t->5.978887138}, {t->11.61730758}, {t->12.13526959}, {t->12.2204946}, {t->17.93561673} }



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