Solve function returns polynomial

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tyler Wallis
tyler Wallis am 9 Feb. 2018
Kommentiert: Walter Roberson am 9 Feb. 2018
I am trying to use the solve function to solve 2 equations with 2 unknowns (a whole number and an angle). I have already solved this by hand but I would like to figure out what I am doing wrong here for future problems and my own sanity. Any help would be appreciated! The answers are FXT_1=144 and Theta_1=38.3
syms FXT theta
F1T=1280;
F2T=50;
F6=70;
v12=0.27;
v21=0.06;
E1=45000;
E2=11000;
G12=4500;
e1T=F1T/E1;
gam6=F6/G12;
sigma1=FXT*cosd(theta)^2;
sigma2=FXT*sind(theta)^2;
sigma6=FXT*sind(theta)^2*cosd(theta)^2;
eqn1=(1/E1)*(sigma1-v12*sigma2)==e1T;
eqn2=-sigma6/G12==gam6;
sol=solve([eqn1,eqn2],[FXT,theta])
FXT_1=sol.FXT
Theta_1=sol.theta
  2 Kommentare
Walter Roberson
Walter Roberson am 9 Feb. 2018
Bearbeitet: Walter Roberson am 9 Feb. 2018
FXT = -5074.671595, theta = 83.20726294
FXT = -5074.671595, theta = -83.20726294
FXT = 1195.771595, theta = 13.37182242*I
FXT = 1195.771595, theta = -13.37182242*I
FXT = -5074.671595, theta = 96.79273706
FXT = -5074.671595, theta = -96.79273706
FXT = 1195.771595, theta = 180.0000000-13.37182242*I
FXT = 1195.771595, theta = 180.0000000+13.37182242*I
and additional solutions for each theta value, every integer multiple of 360*pi .
My testing shows there are certainly no solutions near 144, 38.3
Walter Roberson
Walter Roberson am 9 Feb. 2018
With that sind()^2*cosd()^2 you have to expect that there might be a solution in every quadrant.

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