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From: adel tekari on 22 Mar 2010 12:20
Hi and thank you for your answer,

As I understand, the fzero solves equations with an "original guess" of the solution. And it returns an unique solution, where the f changes the sign.
In my case I have to divide into [-pi 0] and [0 pi].
>> f = @(t) 4*sin(t)-tan(pi/30)*6*cos(t)+tan(pi/30)*4/3;
>> fzero(f,[-pi 0])
ans =
-2.9506
And
>> fzero(f,[0 pi])
ans =
0.1218

"Matt Fig" <spamanon(a)yahoo.com> wrote in message <ho83fi$2lu$1(a)fred.mathworks.com>...
> For numeric values, you could also use FZERO.
>
> f = @(t) 4*sin(t)-tan(pi/30)*6*cos(t)+tan(pi/30)*4/3;
> fzero(f,.1218) % See the help for FZERO.
From: Torsten Hennig on 22 Mar 2010 04:34
> Dear Matlab users
>
> I'm trying to solve the following equation:
>
> t=(0:pi/30:2*pi);
> A=solve('4*sin(t)=tan(pi/30)*6*cos(t)-tan(pi/30)*4/3',
> 't');
>
> I just need the numerical (explicit) values of A. I
> want to see the t=0.1218 and t=-2.9506
> Thanks

Square your equation and substitute
sin^2 t = 1 - cos^2 t
to get a quadratic equation in
cos t (which can easily be solved analytically).

Best wishes
Torsten.
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