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I have this equation in terms of x0 and I need to find x0 in terms of Theta .

theta[x0_] := ArcSin[ 2*(157/100)*(10^(-11))*x0*((1 - 1/x0)^(-1/2))]. 

What is the best way to do that on Mathematica? I've tried to use findroot for x0>3/2 but I get errors and I don't get accurate values. I understand that I should maybe use Nsolve but I'm not sure what's the best way to approach such a problem.

x0f[N_] := x0 /. FindRoot[theta[x0] == N, {x0, 2}] 

Appreciate any suggestions.

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    $\begingroup$ Try: Solve[the == theta[phi], phi] $\endgroup$ Mar 9 at 8:34

1 Answer 1

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We can use inverse function as follows

theta[x0_] := 
 ArcSin[2*(157/100)*(10^(-11))*x0*((1 - 1/x0)^(-1/2))]

t = 
 InverseFunction[ArcSin[2*(157/100)*(10^(-11))*#*((1 - 1/#)^(-1/2))] &]

(*Out[]= 50000000/
  157 ((1000000000 (10/3)^(1/3)
      Sin[#1]^2)/(-1413 Sin[#1]^2 + 
      Sqrt[3] Sqrt[
       665523 Sin[#1]^4 - 100000000000000000000000000 Sin[#1]^6])^(
    1/3) + (10/3)^(
     2/3) (-1413 Sin[#1]^2 + 
       Sqrt[3] Sqrt[
        665523 Sin[#1]^4 - 100000000000000000000000000 Sin[#1]^6])^(
     1/3)) &*)

Visualization

Plot[theta[x], {x, 3/2, 10}, PlotRange -> All, 
 AxesLabel -> {"x0", "theta"}]

Plot[t[x], {x, theta[3/2], theta[10]}, AxesLabel -> {"theta", "x0"}]

Figure 1

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