I'm not sure what the trial-and-error method is for this, but I think that basic Algebraic methods can be used to rearrange the equation. I followed the example in fzero and make an assumption:
% Assume
gamma = 1.55526564459735;
% function to be solved
f = @(M) (1/M)*((2*(1 + (gamma - 1)/2*M^2))/(gamma + 1))^((gamma + 1)/(2*(gamma - 1))) - Ax/Astar;
% initial guess
M0 = 3;
% root-finding solver
Msol = fzero(f, M0)
There should be two real roots.
