# How to find points of intersections of graphs?

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Asatur Khurshudyan on 15 Dec 2012
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?
##### 2 CommentsShowHide 1 older comment
Asatur Khurshudyan on 15 Dec 2012
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.

Azzi Abdelmalek on 15 Dec 2012
Edited: Azzi Abdelmalek on 15 Dec 2012
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
Azzi Abdelmalek on 15 Dec 2012
tzeros{1} is the vector time result for k=1
tzeros{2} is the vector time result for k=2
and so on