# Plotting & Solving

I'm trying to plot replicate a graph from one of the articles and it has series of equations it has solved.

here is my code so far:

Tan [o] == 2 Cot[B]*(M^2*Sin[B]*Sin[B] - 1)/(M^2*(1.4 + Cos[2 B]) + 2);

Pr1[o_, M_] := 1 + 2*1.4/2.4*(M^2*Sin[B]*Sin[B] - 1)
Plot[Table[Pr1[o, M], {o, {10*Pi/180}}] // Evaluate, {M, 1, 3}, AxesOrigin -> {1, 0}]


The variables o, B and M are related to the tangent expression. After that I'm trying to solve the next equation and plot that with certain value of o and as M varies from 1 to 3. For some reason, I'm not getting any error but my graph isn't showing up at all.

Also note that the tan expression produces 6 roots for given M and o, I want to only take the lowest positive value out of the 6 roots.

### Edit

Here is my new code. I used the 4th root instead of 1st one. Thank to belisarius for helping out:

s = Solve[Tan[o] == 2 Cot[B]*(M^2*Sin[B]*Sin[B] - 1)/(M^2*(1.4 + Cos[2 B]) + 2), {B}];

Pr1[o_, M_] := 1 + 2*1.4/2.4*(M^2*Sin[B]*Sin[B] - 1)
Plot[Table[Evaluate[Pr1[o, M] /. s[[4]]], {o, {10*Pi/180}}], {M, 1, 3},
AxesOrigin -> {1, 0}, Evaluated -> True]


The plot shows a step near the value of ~1.4 of M. I want to eliminate this. Any suggestions?

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You can either take the region after that point with the option PlotRange -> {{1.425, 3}, All} or exclude that region with the option RegionFunction -> Function[{x, y}, (x > 1.425)]. – Spawn1701D Apr 12 '13 at 3:52
For Information: Using single capital letters for variable names is discouraged as they often conflict with built in functions, i.e. D, N. – image_doctor Apr 12 '13 at 11:00

A kickstart:

s = Solve[ Tan[o] == 2 Cot[B]*(M^2*Sin[B]*Sin[B] - 1)/(M^2*(1.4 + Cos[2 B]) + 2), {B}];

Pr1[o_, M_] := 1 + 2*1.4/2.4*(M^2*Sin[B] - 1)
Plot[Table[Pr1[o, M] /. s[[1]], {o, 0, 2 Pi, Pi/4}], {M, 1, 3},
AxesOrigin -> {1, 0}, Evaluated -> True]


You have to be careful about what solutions you use from s

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Thank you! I just realized I made few corrections which are edited in the question now. I'm using the 4th root which is the first positive one. Also, the new graph has a sort of a step function before M = 1.4, any way for me to eliminate from the graph? – user6850 Apr 11 '13 at 19:33
Why use bot Evaluate and Evaluated -> True? – m_goldberg Apr 13 '13 at 7:29
@m_goldberg I was trying both and forgot to remove one of them when posting. Thanks. – Dr. belisarius Apr 13 '13 at 11:04