You are essentially trying to solve a system of two equations, at least in general. For the simple case where one of the equations is a constant, thus y = 1, we can solve it using fzero. Of course, it is always a good idea to use graphical means to find a good starting point.
f=@(t) (9*(sin(t))./t) + cos(t);
y0 = 1;
The idea is if you want to find where the two curves intersect, is to subtract them, then look for a root of the resulting difference.
(By the way, note that I used ./ for the divide, so that MATLAB won't have problem for vector or array input in f. This is a good habit to develop.)
Note that f(t) is not strictly defined in MATLAB at zero, since it results in 0/0. (A limit exists of course for the function, and can be evaluated using my limest tool.)
limest(f,0)
ans =
10
Since I know the solution is not at 0, I'll just use the fzero bounds from looking there for a root.
format long g
fzero(@(t) f(t) - y0,[eps,6*pi])
ans =
2.58268206208857
But is this the only root? What if we have two or more solutions? Finding all the roots of a completely general function can be a nasty problem, as some roots may be infinitely close together, or there may be infinitely many roots.
One idea is to use a tool that knows how to look for multiple solutions to a problem. Again, found on the file exchange, we can use research.
y0 = 1;
rmsearch(@(t) f(t) - y0,'fzero',1,eps,6*pi)
ans =
2.58268206208857
6.28318530717959
7.97464518075547
12.5663706143592
13.7270312712311
y0 = -1;
rmsearch(@(t) f(t) - y0,'fzero',1,eps,6*pi)
ans =
3.14159265358979
5.23030501095915
9.42477796076938
10.8130654321854
15.707963267949
16.6967239156574