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I have the next equation:

enter image description here

I wrote it in mathematica and I obtained

f[h_] = 8 - 5 (-(2 - h) Sqrt[4 h - h^2] + 4 Sec[(2 - h)/2])

Now I have to solve it, finding the h. I know that h is between 0<=h<=4 by the Sqrt equation but I don't know how to find the real root. I tried with NSolve or FindRoot but it didn't work.

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    $\begingroup$ you know the notation cos^-1 usually denotes inverse cosine not 1/cos..(sec) $\endgroup$
    – george2079
    Apr 25, 2018 at 1:21
  • $\begingroup$ indeed changing Sec to ArcCos you should get a solution around h=.74 $\endgroup$
    – george2079
    Apr 25, 2018 at 1:29

1 Answer 1

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EDIT: Corrected eqn for missing L 

eqn = V == (r^2 ArcCos[(r - h)/r] - (r - h) Sqrt[2 r h - h^2])L;

const = {r -> 2, L -> 5, V -> 8};

The exact solution for h is a Root object

sol = Solve[eqn /. const, h, Reals][[1]]

{h -> Root[{-(8/5) + 4 ArcCos[(2 - #1)/2] - 
      2 Sqrt[-(-4 + #1) #1] + #1 Sqrt[-(-4 + #1) #1] &, 
    0.74001521805594051394}]}

The numeric approximation is

sol // N

(* {h -> 0.740015} *)

Verifying that the solution satisfies the equation

eqn /. const /. sol // FullSimplify

(* True *)

or

eqn /. const /. N[sol]

(* True *)

Plot[Evaluate[eqn /. const], {h, 0, 4},
 Epilog -> {Red, PointSize[Medium], Point[{h /. sol, 0}]}]

enter image description here

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