MATLAB: Multiple Answers From ‘solve’

functionssolve

I have an equation that I am passing to the 'solve' function. I know for a fact that it has two solutions. It only returns one solution. This is how I'm calling solve:

solve(S)

where S is a string containing the equation. It's pretty ugly but I can provide it if needed. What I need is a way to get that second solution.

Best Answer

syms theta2 theta4
S=(-4.19*sin(theta2*pi/180)-5.166*sin(theta4*pi/180)+23.3*sin(82.99*pi/180))^2+(-4.19*cos(theta2*pi/180)-5.166*cos(theta4*pi/180)+23.3*cos(82.99*pi/180))^2-24.073^2;
sols = solve(S,theta4);
sols = subs(sols,theta2,73.5);
   sols =
                                [-115.9829194]
                                [            ]
                                [-73.90876566]

The first solution, added to 180, gives 64.0170806.

Related Solutions

MATLAB: How to solve a non-linear equation system as function handle with fsolve

qini = zeros(1,6);             % Initial guess
ci = fsolve(@funec,qini)       % fsolve with the function handle, unknown variables and initial guess
                               % done many variations here, none of them worked
function ec = funec(x);
q1 = x(1);
q2 = x(2);
q3 = x(3);
q4 = x(4);
q5 = x(5);
q6 = x(6);
ec =[...
    sin(q6)*(cos(q1)*cos(q2)*sin(q3)*sin(q4) - cos(q4)*sin(q1) + cos(q1)*cos(q3)*sin(q2)*sin(q4)) - cos(q6)*(cos(q2 + q3)*cos(q1)*sin(q5) + cos(q5)*sin(q1)*sin(q4) + cos(q1)*cos(q2)*cos(q4)*cos(q5)*sin(q3) + cos(q1)*cos(q3)*cos(q4)*cos(q5)*sin(q2)) - 3097002709122497/18014398509481984
    cos(q6)*(cos(q1)*cos(q4) + cos(q2)*sin(q1)*sin(q3)*sin(q4) + cos(q3)*sin(q1)*sin(q2)*sin(q4)) + sin(q6)*(cos(q2 + q3)*sin(q1)*sin(q5) - cos(q1)*cos(q5)*sin(q4) + cos(q2)*cos(q4)*cos(q5)*sin(q1)*sin(q3) + cos(q3)*cos(q4)*cos(q5)*sin(q1)*sin(q2)) - 6124274932042799/36028797018963968
    - sin(q2 + q3)*cos(q5) - cos(q2 + q3)*cos(q4)*sin(q5) - 139701203459709/140737488355328
    70*cos(q1) + 360*cos(q1)*sin(q2) - 65*sin(q1)*sin(q4)*sin(q5) + 380*cos(q1)*cos(q2)*cos(q3) - 380*cos(q1)*sin(q2)*sin(q3) + 65*cos(q1)*cos(q2)*cos(q3)*cos(q5) - 65*cos(q1)*cos(q5)*sin(q2)*sin(q3) - 65*cos(q1)*cos(q2)*cos(q4)*sin(q3)*sin(q5) - 65*cos(q1)*cos(q3)*cos(q4)*sin(q2)*sin(q5) + 19869/50
    sin(q6)*(cos(q1)*cos(q4) + cos(q2)*sin(q1)*sin(q3)*sin(q4) + cos(q3)*sin(q1)*sin(q2)*sin(q4)) - cos(q6)*(cos(q2 + q3)*sin(q1)*sin(q5) - cos(q1)*cos(q5)*sin(q4) + cos(q2)*cos(q4)*cos(q5)*sin(q1)*sin(q3) + cos(q3)*cos(q4)*cos(q5)*sin(q1)*sin(q2)) + 2202239090516653/2251799813685248
    cos(q2 + q3)*cos(q6)*sin(q4) - sin(q6)*(sin(q2 + q3)*sin(q5) - cos(q2 + q3)*cos(q4)*cos(q5)) + 7591943296130745/288230376151711744
    cos(q2 + q3)*cos(q1)*cos(q5) - sin(q5)*(sin(q1)*sin(q4) + cos(q1)*cos(q2)*cos(q4)*sin(q3) + cos(q1)*cos(q3)*cos(q4)*sin(q2)) - 102014544375725/18014398509481984
    70*sin(q1) + 360*sin(q1)*sin(q2) - 380*sin(q1)*sin(q2)*sin(q3) + 380*cos(q2)*cos(q3)*sin(q1) + 65*cos(q1)*sin(q4)*sin(q5) + 65*cos(q2)*cos(q3)*cos(q5)*sin(q1) - 65*cos(q5)*sin(q1)*sin(q2)*sin(q3) - 65*cos(q2)*cos(q4)*sin(q1)*sin(q3)*sin(q5) - 65*cos(q3)*cos(q4)*sin(q1)*sin(q2)*sin(q5) - 1563/20
    cos(q6)*(sin(q2 + q3)*sin(q5) - cos(q2 + q3)*cos(q4)*cos(q5)) + cos(q2 + q3)*sin(q4)*sin(q6) - 2129854249224409/18014398509481984
    cos(q6)*(cos(q1)*cos(q2)*sin(q3)*sin(q4) - cos(q4)*sin(q1) + cos(q1)*cos(q3)*sin(q2)*sin(q4)) + sin(q6)*(cos(q2 + q3)*cos(q1)*sin(q5) + cos(q5)*sin(q1)*sin(q4) + cos(q1)*cos(q2)*cos(q4)*cos(q5)*sin(q3) + cos(q1)*cos(q3)*cos(q4)*cos(q5)*sin(q2)) - 8872946535830847/9007199254740992
    cos(q2 + q3)*cos(q5)*sin(q1) - sin(q5)*(cos(q2)*cos(q4)*sin(q1)*sin(q3) - cos(q1)*sin(q4) + cos(q3)*cos(q4)*sin(q1)*sin(q2)) - 8718732974408059/72057594037927936
    360*cos(q2) - 380*cos(q2)*sin(q3) - 380*cos(q3)*sin(q2) - 65*cos(q2)*cos(q5)*sin(q3) - 65*cos(q3)*cos(q5)*sin(q2) - 65*cos(q2)*cos(q3)*cos(q4)*sin(q5) + 65*cos(q4)*sin(q2)*sin(q3)*sin(q5) - 39571/100];
end

MATLAB: Solving non linear equation using fsolve

You are not passing theta_n and other parameters to root2d(). One fix is to make root2d a Nested Function. Also, you must choose a different initial point. phi0 = [10,10] worked for me.

function solve_it()
    beta_a = 10;
    alpha_n = 10 ;
    eta= 18;
    i = 18;
    theta_i = 10;
    theta_n = 10 ;
    
    fun = @root2d;
    phi0 = [10,10];
    phi = fsolve(fun,phi0);
    phi_n = phi(1),
    phi_i = phi(2),
    function F = root2d(phi)
    F = zeros(2,1);
    F(1) = (pi*sin((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2) - (pi*cos((pi*phi(1))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((phi(1)*pi)/180)^2*(cos((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180) + sin((pi*phi(2))/180)*tan((pi*theta_i)/180)));
    F(2) = ((cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180))*((pi*cos((pi*(phi(1) + theta_n))/180)*sin((pi*phi(2))/180))/180 - (pi*cos((pi*phi(2))/180)*tan((pi*theta_i)/180))/180))/(sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2);
    end
end

Best Answer

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