how can we solve non-linear equations like this via MATLAB, I am trying to solve it by using following program but it shows error, shown below in the pic? Thanks in advance.

clc

clear
syms a1 a2 a3 a4
eqn1 = cos(a1)+cos(a2)+ cos(a3)+ cos(a4) ==3
eqn2 =  cos(3*a1)+ cos(a3*2)+ cos(3*a3)+ cos(3*a4) ==0
eqn3 =  cos(5*a1)+ cos(5*a2)+ cos(5*a3)+ cos(5*a4) ==0
eqn4 =    cos(7*a1)+ cos(7*a2)+ cos(7*a3)+ cos(7*a4) ==0
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4], [a1, a2, a3, a4])
X = linsolve(A,B)

[Alex Sha]

There are multi-solutions:

1:

a1: -1.11780969756713
a2: -0.464078788776745
a3: -0.203844647896775
a4: 0.810939988237884

2:

a1: 1.11780971343386
a2: 0.464078821289344
a3: 6.07934066679302
a4: -5.4722453083044

3:

a1: 0.415439221706074
a2: 1.07905297726765
a3: 7.14839217493061
a4: 6.55076057446668

[Muhammad Zain Khan]

how you solved the problem?

source code?

[Walter Roberson]

Alex uses third-party commercial software "1stOp"

Answer 1

[Asad (Mehrzad) Khoddam]

You have used cos function and your equations are not linear.

[Muhammad Zain Khan]

Assalam-o-Alikum Asad! Can you guide me on how to approach the problem?

Edited: thanks I solved it by myself, but your mentioning of cosine, that it is not linear was really helpful. Thank you.

Answer 2

[Asad (Mehrzad) Khoddam]

eqns = [2*u + v == 0, u - v == 1];
S = solve(eqns,[u v])
clear
syms a1 a2 a3 a4
eqn1 = cos(a1)+cos(a2)+ cos(a3)+ cos(a4) ==3
eqn2 =  cos(3*a1)+ cos(a3*2)+ cos(3*a3)+ cos(3*a4) ==0
eqn3 =  cos(5*a1)+ cos(5*a2)+ cos(5*a3)+ cos(5*a4) ==0
eqn4 =    cos(7*a1)+ cos(7*a2)+ cos(7*a3)+ cos(7*a4) ==0
eqns = [eqn1, eqn2, eqn3, eqn4];
s = solve(eqns,[a1, a2, a3, a4])

[Muhammad Zain Khan]

The solution you suggested is taking too long to evalute the values or in my case it is not even evaluating. I used this approach to solve the problem: But it gives variable values for 3 equations, (1,2,4) and for 3rd equation it have some non-zero answer.

% MAPPING: b(1) = a1,  b(2) = a2,  b(3) = a3,  b(4) = a4
f = @(b) [cos(b(1))+cos(b(2))+ cos(b(3))+ cos(b(4))-3
    cos(3*b(1))+ cos(3*b(2))+ cos(3*b(3))+ cos(3*b(4))
    cos(5*b(1))+ cos(5*b(2))+ cos(5*b(3))+ cos(5*b(4))
      cos(7*b(1))+ cos(7*b(2))+ cos(7*b(3))+ cos(7*b(4))];
B0 = [0.4090; 0.1893; 4.6979;3.4790]*2*pi;
[B,fv,xf,ops] = fsolve(f, B0);
ps = ['a1'; 'a2'; 'a3'; 'a4'];
fprintf(1, '\n\tParameters:\n')
for k1 = 1:length(B)
    fprintf(1, '\t\t%s = % .4f\n', ps(k1,:), B(k1))
end

[Asad (Mehrzad) Khoddam]

change

 cos(5*b(1))
 to 
  cos(5*b(4))  % third equation
  

in your code. This approach is nemerical solution approach.