How to find points of intersections of graphs?

Hi everybody. I need to find zeros of given functions of one (and the same) variable in given interval. f_k(t)=1-cos(a_k*t-b_k), -Pi<t<Pi, k=1,2,...,n (n is a given integer), a_k and b_k are known.

I thought to draw graphics of that functions and find the points of their intersections. But I could just draw graphics, and could not find that points. Could you help me with graphics or advice another way of solution?

[Matt Fig]

The question is a little unclear. Do you mean that for each integer k on [1,n] you want to find where 1-cos(a(k)*t-b(k)) intersects all of the other 1-cos(a(d)*t-b(d)), d~=k, d on [1,n]?

How many long is a and b? Can you show them?

[Asatur Khurshudyan]

Yes, you`re right. I can right down only a_k here: a_k=(Pi*k)^2/sqrt(1+c*(Pi*k)^2), where c is given constant. b_k are of the arctan() form.

Answer 1

[Azzi Abdelmalek]

pas=0.01
t=-pi:pas:pi;
n=length(t);
a_k=[1 2 3 4];
b_k=[1 2 3 4];
for k=1:numel(a_k)
f_k{k}=1-cos(a_k(k)*t-b_k(k));
[c,idx1]=findpeaks(-f_k{k})
tzeros{k}=t(idx1)
end

[Asatur Khurshudyan]

Thank you for response. But could you, please, explain your answer?

[Azzi Abdelmalek]

Ok,I've noticed that the zeros of your function f_k corresponds to its minima. findpeaks function will find the maxima of your function f_k, then I used findpeaks(-f_k) to find a minimum.

 [c,idx]=findpeaks(-f) 
 % c are the minimum
 % idx are indexes corresponding to each minimum

[Matt Fig]

This Answer does not match at all with what you said above in the comments. It finds the minimum (zero) of the given function for each k individually. Above you said you meant something entirely different! Perhaps there is a communication problem.

[Asatur Khurshudyan]

You are right, I don`t need solution for EACH k (it is very obvious: t_j=(2*Pi*j+b_k)/a_k), I need the solution of the one-parametric SYSTEM 1-cos(a_k*t-b_k)=0, k=1,2,...,n in -Pi<t<Pi interval, which should satisfy all that equations simultaneously. I think, that graphical method suits the best.

[Azzi Abdelmalek]

you can choose the intersection of vectors in tzeros. In your case it's t0=0.9984

[Asatur Khurshudyan]

Haw can I check your algorithm (the problem is that I`m just a beginner)?

[Azzi Abdelmalek]

tzeros{1} is the vector time result for k=1

tzeros{2} is the vector time result for k=2

and so on