Roots of a function
7 views (last 30 days)
Show older comments
I have two equations:
y1 = 2sinx1;
y2 = 2cos^2(x1) + 3sin(2x2+3);
here y1 = 0 while y2 = 1.
Can anyone please tell me which approach would be the best to find out the values of x1 and x2.
Thank you.
Accepted Answer
Torsten
on 31 Mar 2016
Edited: Torsten
on 31 Mar 2016
Try this:
function driver
y=zeros(1,2);
y(1)=0;
y(2)=1;
x0=zeros(1,2);
x=fsolve(@(x)HW1(x,y),x0)
function res = HW1(x,y)
res(1)=y(1)-2*sin(x(1));
res(2)=y(2)-(2*cos(x(1))^2 + 3*sin(2*x(2) + 3));
Best wishes
Torsten.
2 Comments
Torsten
on 31 Mar 2016
Your equations don't have a unique solution. So - depending on the starting guess x0 - you'll get different solutions for x. Do you have any condition on x that could make the solution unique ?
Best wishes
Torsten.
More Answers (1)
Vlad Miloserdov
on 30 Mar 2016
if you still need this
A=solve('0 = 2*sin(x1)','1 = 2*cos(x1)^2 + 3*sin(2*x2+3)','x1','x2');
% first ans
A.x1(1)
A.x2(1)
% second ans
A.x1(2)
A.x2(2)
5 Comments
See Also
Categories
Find more on Function Creation in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!