Solving a system of equation in terms of an unknown.

Hi, I'm new here and quite new in MatLab too. So I have a question about an equation sistem consisting of 4 equations and having 5 (tet1, tet2, tet3, x, q) unknowns.. I'm trying to solve 4 of these unknowns (tet1, tet2, tet3, x) in terms of the other one (q). I've written a code, it returns an error in like 30 mins.

Here is the code:

clear
close all
clc

alp = 150;
bet = 18;
a = 0.0125;
b = 0.035;
c = 0.0147;
d = 0.043;
h = 0.0195;
m = 0.0113;
n = 0.0344;

syms x tet1 tet2 tet3 q;

eq(1) = m + d*cos(tet1+bet) - b*cos(tet3) - a*cos(q) == 0;
eq(2) = n + d*sin(tet1+bet) - b*sin(tet3) - a*sin(q) == 0;
eq(3) = x*cos(tet1) + h*cos(tet2) - c*cos(180-alp+tet3) - d*cos(tet1+bet) == 0;
eq(4) = x*sin(tet1) + h*sin(tet2) - c*sin(180-alp+tet3) + d*sin(tet1+bet) == 0;

Sol1 = solve(eq,[x tet1 tet2 tet3]);

What am I missing? What should I change in the code. Thank you in advance for the help.

[Walter Roberson]

Using Maple, I am able to get solutions in terms of q, by solving for x, tet1, tet3, then tet2 . There are two basic solution families. However, the expressions for tet2 in terms of q are over 19000 characters in each of the families.

Note that you are trying to derive exact solutions to equations involving floating point coefficients. That is not justifiable. For example, in science, c = 0.0147 means that c is some value in the range [147/1000 - 5/10000, 147/1000 + 5/10000) -- and if you do not know the exact value of c, then how could you justify a 20 Kb long exact formula involving some nominal value of c in that range ?

Answer 1

[Ameer Hamza]

Your system of equations is very nonlinear, involving several trigonometric functions, so it is unlikely that it has a closed-form solution. You can solve it for a particular value of 'q' and get a numeric solution. For example, this code solves the equations for q=0.5

clear
close all
clc
alp = 150;
bet = 18;
a = 0.0125;
b = 0.035;
c = 0.0147;
d = 0.043;
h = 0.0195;
m = 0.0113;
n = 0.0344;
syms x tet1 tet2 tet3 q;
eq(1) = m + d*cos(tet1+bet) - b*cos(tet3) - a*cos(q) == 0;
eq(2) = n + d*sin(tet1+bet) - b*sin(tet3) - a*sin(q) == 0;
eq(3) = x*cos(tet1) + h*cos(tet2) - c*cos(180-alp+tet3) - d*cos(tet1+bet) == 0;
eq(4) = x*sin(tet1) + h*sin(tet2) - c*sin(180-alp+tet3) + d*sin(tet1+bet) == 0;
Sol1 = vpasolve(subs(eq, q, 0.5),[x tet1 tet2 tet3]);

Answer 2

[Mustafa Cüneyt Sevinç]

I guess I'll stick to your solving method Ameer Hamza since I have the "q" values from 0 to 360 dergrees by 10 degrees steps at a time, making a total of 37 "q" values. Thank you for your answers guys. Appreciate it so much!