How to solve a system of nonlinear equations with three unknowns & three equations?

I tried to solve the system of nonlinear equations with fsolve function but the answers for unknowns are not true,I think there is a problem with number of iterations or the reasonable error.

How can I set these parameters?

Here are the equations:

311*cos(0.5236-x(3))+x(1)*cos(x(2)+x(3))-311=0
311*sin(.5236-x(3))-(x(1)*sin(x(2)+x(3)))=0
(x(1)^2)*(tan(.5236)*cos(x(2))*cos(x(2)-.7526)-(1/tan(.5236))*(sin(x(2))*sin(x(2)-.7526)))-7036.18=0

x(1) is a positive and x(2) and x(3) are angles then it is reasonable to be between 0 and 2pi

[Azzi Abdelmalek]

Show your equations

[Farid]

Here are my equations: 311*cos(0.5236-x(3))+x(1)*cos(x(2)+x(3))-311 & 311*sin(.5236-x(3))-(x(1)*sin(x(2)+x(3))) & (x(1)^2)*(tan(.5236)*cos(x(2))*cos(x(2)-.7526)-(1/tan(.5236))*(sin(x(2))*sin(x(2)-.7526)))-7036.18

[sixwwwwww]

What are the conditions for your equations? are these equations equal to each other?

[Walter Roberson]

Probably that all of them should equal 0.

[Farid]

Yeah, all of them are equal to zero, x(1) is a positive and x(2) and x(3) are angles then it is reasonable to be between 0 and 2pi

Answer 1

[Matt J]

Seems a matter of making a lucky initial guess. Here's one solution

x =

    146.6872    3.9507    0.9997

Here's another

x =

    592.2262    2.9287    9.3271

And another

x =

    592.2262    2.9287    3.0439

[Farid]

can you tell me about the code which generated these answers or send it to me? (<mailto:[e-mail] [e-mail]>)

[Matt J]

All I did was experiment with different initial points

 x0=[311,2,2];
 x0=[650 3 5];

[Farid]

ok, with which command then?

[Matt J]

Both with fsolve and lsqnonlin. Note that lsqnonlin allows you to apply bounds which is good for x(2) and x(3). The function is 2*pi periodic in x(2) and x(3), so bounds allows the solver to ignore periodic repetitions.

[Farid]

Can the command Isqnonlin be used in simulation? I am sure that tha command fsolve can not be used in simulink I mean for example you can not use it in a ambedded MATLAb function.

[Matt J]

That, I do not know.

[Farid]

ok,thanks about your consideration

Answer 2

[Walter Roberson]

x = [146.6871682461996007408428, 3.950714907905676528151964, 0.9997481449580266590840019]

To the accuracy used, this is the only solution with x(1) positive and the other two in [0, 2*Pi]

The x(1) in the 592 range (given by Matt J) appear to be false solutions.

Method of solution:

The third equation can be solved for x1 to give x1 in terms of x2.

The second equation can be solved for x3 to give x3 in terms of x2.

Substituting the x1 and x3 terms into the first equation gives an equation in terms of x2 only. This equation goes imaginary for some values of x2 in [0, 2*Pi] . By examining the plot, you can see that there is a single 0 crossing, somewhere near x2 = 5*Pi/4. You can use fzero() or equivalent to find the x2 for which the equation becomes 0.

Then, with the numeric x2 in hand, you can substitute into the two previous forms to get x1 and x3. Although there is an arctan() involved, there is only one solution that works.

[Matt J]

The second equation can be solved for x3 to give x3 in terms of x2.

Can't immediately see how you did this. The second equation involves all 3 variables, right? When substituting x1 as a function of x2 into the 2nd equation, you end up with something pretty hideous. You say it can be solved for x3 regardless?

In any case, here's the solution with x1 near 592 to more precision. You can substitute into the equations directly to verify that it is a solution.

x =

            592.226174086239
            2.92865236277986
            3.04387323520966

[Walter Roberson]

Ummm... now I'm scratching my head about how I got those results. Maybe when I wake up in the morning.

[Farid]

these equations are not completely true and maybe the some coefficients should be changed,I'm looking for a general solution for these equation in MATLAB codes

[Walter Roberson]

Your equations involve the values 0.5236 and 0.7526 . Is there a formula that relates those two values? If I replace 0.5236 with the symbol alpha, then can 0.7526 be expressed in terms of alpha, or is it an independent variable?