what's the relation between A , B and M,G for this Nonlinear system of equation ?

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A*sin(3*Phi)-B*sin(Phi) = G ;
A*cos(3*Phi)-B*cos(Phi) = M ;

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Torsten il 11 Feb 2024
A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2
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Torsten il 11 Feb 2024
Modificato: Torsten il 11 Feb 2024
If each of the equations could be solved uniquely for Phi, you could get what you want.
Assume that the first equation would uniquely yield Phi = f(A,B,G) and the second equation would uniquely yield Phi = g(A,B,M), then f(A,B,G) - g(A,B,M) = 0 would be your relation. But unfortunately, the equations cannot be solved uniquely for Phi.
syms A B G M Phi
eqn = A*sin(3*Phi)-B*sin(Phi) == G ;
sol1 = solve(eqn,Phi,'ReturnConditions',1,'MaxDegree',3)
sol1 = struct with fields:
Phi: [6×1 sym] parameters: k conditions: [6×1 sym]
ans = 
ans = 

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John D'Errico
John D'Errico il 11 Feb 2024
Modificato: John D'Errico il 11 Feb 2024
This is not even remotely a question about MATLAB. As such, it should arguably not even be on Answers. But I have a minute to respond, so I will choose to do so.
Trivial! What is the relation? Admittedly, the relation itself is a slightly complex thing, composed of two equations. The relations are:
A*sin(3*Phi)-B*sin(Phi) = G
A*cos(3*Phi)-B*cos(Phi) = M
which is exactly what you wrote.
It is not a nonlinear system of equations though. Not at all! Phi there is simply a parameter, not one of the parameters involved. That makes your problem fully a LINEAR system of equations. As such, if you want to view it in that form, then we could write:
M = [sin(3*Phi), -sin(Phi); ..
cos(3*Phi), -cos(Phi)]
So M is a matrix function of the parameter Phi. Biven the matrix M, then we could write:
M*[A;B] = [G;M]
There is no simpler relation between those variables. And it is NOT at all nonlinear. Purely linear.
  1 Commento
YOUSSEF El MOUSSATI il 11 Feb 2024
Ok ,thanks for your answer , yes this system is linear but the parameters A,B and C,D are Nonlinear A=f(x^3,x^2,x); B=g(x^3,x^2,x); C=h(x^3,x^2,x); D=l(x^3,x^2,x);

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