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Is there a way to get the analytical form to the root of this equation:

2 A e (-ArcTanh[smax/s0] + ArcTanh[(smax Cos[w ζ])/s0]) == F0

ζ is the variable to be solved and is bounded by (1/2 π / w, π / w).

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Well ... Solve[2 A e (-ArcTanh[smax/s0] + ArcTanh[(smax Cos[w x])/s0]) == F0, x] solves it. What else do you need? – Dr. belisarius Feb 10 '13 at 5:16
up vote 5 down vote accepted

Your form of equation is overpopulated with redundant constants slowing down computation. Rewrite and solve to get a simple concise form:

Solve[Pi/2 < x < Pi && ArcTanh[b Cos[x]] == a, x, Reals]

enter image description here

Get back to your original form:

% /. {a -> F0/(2 A e) + ArcTanh[smax/s0], b -> smax/s0, x -> w \[Zeta]}
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thanks, Vitaliy Kaurov and Nasser. :) – xinxin guo Feb 10 '13 at 10:14

@belisarius advice is good. But for someone who is new to Mathematica, then they should take advantage of the new suggestion bar in V9.

These are the steps to apply if you do not know the commands to use. By just typing the same line you showed, you will obtain the suggestion bar. By clicking on more you will see the command to solve for $\zeta$ there. Clicking on it will generate the result. The following diagram illustrates these steps taking full advantage of the new and powerful suggestion bar in V9.

enter image description here

In addition, Mathematica displays the command used and the solution automatically

enter image description here

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