Epsilon close solution to a trigonometrical system of equations.

I am having the following trigonometrical system of equations that I want to solve, further restricting each angle in a specific interval (-pi/3,pi/3), (-2pi/3,0), (-2pi/3,0) respectively.

syms u1 u2 u3;
eqn = 3.98*cos(u1)+ 2.24*cos(u1+u2)+1.58*cos(u1+u2+u3) == d1;
eqn2 = 3.98*sin(u1)+ 2.24*sin(u1+u2)+1.58*sin(u1+u2+u3) == d2;

[solx, soly, solz, param, cond] = solve([eqn, eqn2], [u1,u2,u3], 'Real', true, 'ReturnConditions', true) 
assume(cond)
solk = solve([solx >= -pi/3, solx <= pi/3, soly<=0, soly>= -2*pi/3,solz <=0,solz>=-2*pi/3],param)

Where u1,u2,u3 are the unknown angles and d1, d2 are the input parameters of my function. The problem are the following: a)Is my use of solve the appropriate? The function needs a lot of time to finish its execution.

b) Apart from the obvious solution (0,0,0) when I give as input (3.98+2.24+1.58,0), in every other input that I tried, I get that there is no solution. I guess this happens due to the high precision needed. For my function to produce an answer I need to find some d1,d2 :

d1^2+d2^2 = 7.8^2 ,

with finitely many decimal digits that is not what I intend. How can I put some e (epsilon) and express the notion that it is enough for the above sum to be epsilon close to

7.8^2

Answer 1

[Roger Stafford]

The problem here is that you have more unknowns than equations. In such a case there is generally an infinite continuum of possible solutions. Matlab does not know which one you want.

[Ioannis Nemparis]

Would adding ('PrincipalValue', true) solve that issue? But what about the other problem?