When you have four non-linear equations with three unknowns you cannot, in general, expect to get a symbolic solution. For these types of problems you will have to resort to some kind of minimization/optimization approach. The first stab to try is typically to do a least-squares minimization of the square of the residuals - if there happens to be one exact solution that will also be a valid solution of the four original equations, if there is some global minima then that is as good as you can do. For this write a function for the square of the residuals, something like this:
function err = you4nonlsqerrfcn(pars,A1,B1,A2,B2,wg,wp)
T1 = pars(1);
T2 = pars(2);
alpha = pars(3);
E1 = (wg*( T1 + T2*alpha ) / ( 1 - alpha*wg^2*T1*T2 )) - A1;
E2 = (wg*( alpha*T1 + T2 ) / ( 1 - alpha*wg^2*T1*T2 )) - B1;
E3 = (wp*( T1 + T2*alpha ) / ( 1 - alpha*wp^2*T1*T2 )) - A2;
E4 = (wp*( alpha*T1 + T2 ) / ( 1 - alpha*wp^2*T1*T2 )) - B2;
err = sum([E1,E2,E3,E4].^2);
end
Then you make a start-guess and try the fminsearch-function:
T1T2alpha0 = [1 2 pi/3];
[T1T2alpha1,Fval,ExFlag] = fminsearch(@(pars) you4nonlsqerrfcn(pars,A1,B1,A2,B2,wg,wp),T1T2alpha0);
If Fval is close enough to zero and ExFlag is 1 you ought to have an acceptable solution, if not you might have to continue the search from T1T2alpha1 or try a new start-guess.
Once you've done this you can also make an attempt to use lsqnonlin with a modified version of the error-function that instead of returning the sum-of-squared of the residuals returns the residuals directly (i.e. E1, E2, E3 and E4). This might be numerically more efficient.
HTH
