How to use fzero in a loop to obtain the first 3 positve solutions for cos3x=sin3x?

As it says in the title, I can't figure a way to do it. I've tried for loops but it didn't quite work.

Thanks!

[dpb]

Well, let's see what you tried...seems straightforward enough although you'll have to give starting values to make fzero find the desired solutions...

Answer 1

[Fabio Freschi]

fzero gives you only one solution. In case of multiple roots, the choice depends on the initial choice and on the algorithm. If you plot cos(3*x)-sin(3*x), you can see that the first 3 solutions are in the interval [0,pi]. So you can loop over x with a small step and then remove the duplicated solutions

% your function
f = @(x)cos(3*x)-sin(3*x);

% picture
figure, hold on
fplot(f,[0,pi]);

% number of divisions
N = 100;

% initial guesses
x0 = linspace(0,pi,N);

% preallocation
xSol = zeros(N,1);

% set tolerance
tol = 1e-4;
options = optimset('TolX',tol);

% now loop over x
for i = 1:N
    xSol(i) = fzero(f,x0(i),options);
end

% filter repeated solutions
xSolUnique = uniquetol(xSol,10*tol);

% filter solution out of bounds
xSolUnique = xSolUnique(xSolUnique > 0 & xSolUnique < pi)
    
% add solutions to the plot
plot(xSolUnique,f(xSolUnique),'o');

[Yuechuan Chen]

Thank you so much! I'm a first year maths student and have only started learning MatLab. I'll study the code to have a better understanding. Thank you!

[Fabio Freschi]

My pleasure!

Basically, I created a set of uniformly distributed points between 0 and pi and sotred them in x0.

The I ran fzero starting from each of these points with the loop. Because some initial starting point may provide the same root, I removed the duplicated solution with uniquetol (the final solutions may differ of the tolerance used by fzero). I finally removed the solution that fell out of the domain 0-pi.

If the solution answer your original problem and you are satisfied, please accept it.

[Yuechuan Chen]

I'm grateful for the help, thank you for the explanation!

Answer 2

[Matt J]

It seems like a strange homework problem, since it is so easily solved analytically. Well, assuming we play dumb to that, you need to use some analysis to find what intervals the first 3 solutions lie in. The equation is equivalent to tan3x=1 and so you know the solutions lie in the open intervals (0,pi/6),(pi/6,3*pi/6), and (3*pi/6,5*pi/6). You can then use fzero to search for roots in these intervals, e.g.,

>> fun=@(x) tan(3*x)-1;

>> x2=fzero(fun,[pi/6+eps,3*pi/6-eps])

x2 =

    1.3090

and similarly for the other solutions.

[Yuechuan Chen]

Thank you very much! Yeah it was one of the questions in my assigment :)