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I have a function and want to re-draw its plot for its three variables. Here is my function:

fig1

Where X is:

fig2

The plot of this function is:

Fig3

I'm thinking to draw a beautiful plot something like this:

Fig4

I was reading this page:

Plots in 2D

But couldn't find any idea to draw it. Is it possible to draw such a plot using Mathematica? How?

Edit1:

Forgot to say if it needed you can assume k=1.4 and M1=2

Edit2:

The code I've tried.

In[1]:= k = 1.4

Out[1]= 1.4

In[2]:= m = 2

Out[2]= 2

In[8]:= X = (Tan[b - thet])/(Tan [b])

Out[8]= Cot[b] Tan[b - thet]

In[9]:= Q = -((k + 1)/2)*X^2*m^2 + (1 + k*m^2) X - (1 + ((k - 1)/2)*m)

Out[9]= -1.4 + 6.6 Cot[b] Tan[b - thet] - 4.8 Cot[b]^2 Tan[b - thet]^2

In[10]:= ContourPlot[Q, {thet, 0, 50}, {b, 0, 90}]
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Clear["Global`*"]

k = 7/5;
m = 2;
X = Tan[b - thet]/Tan[b];
Q = -((k + 1)/2)*X^2*m^2 + (1 + k*m^2) X - (1 + ((k - 1)/2)*m) // FullSimplify

(* 1/5 (-7 + 3 Cot[b] Tan[b - thet] (11 - 8 Cot[b] Tan[b - thet])) *)

Your plot has too much structure to be seen at the PlotRange you used.

ContourPlot[Q, {thet, 0, 5}, {b, 0, 9}, PlotPoints -> 50]

enter image description here

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See example below (omitting eq 19 and 20 from your sample)

q[\[Beta]_, \[Theta]_] := Module[{x = Tan[\[Beta] - \[Theta]]/Tan[\[Beta]]}, 1 - x^2 - 3 x]

ContourPlot[q[\[Beta], \[Theta]], {\[Beta], 0, 1}, {\[Theta], 0, 1}, ContourShading -> None]

screenshot

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