Trigonometric non linear equation
1 回表示 (過去 30 日間)
古いコメントを表示
clc;close all;clear all;
theta=30;
k=360;
h=0.5555;
After running the above program the answer of F iam getting 0.0730
Suppose the value of 'h' is unknown.
So, which function i used to get the value of 'h' where the below equation is equal to zero.
((cosd(theta).*(sind(k*h).^2)) - (2*(sind(k.*h.*cosd(theta)).^2)))==0
0 件のコメント
採用された回答
Star Strider
2022 年 5 月 7 日
Try something like this —
theta=30;
k=360;
h=0.5555;
F = @(h) ((cosd(theta).*(sind(k*h).^2)) - (2*(sind(k.*h.*cosd(theta)).^2)));
for k1 = 1:360
h_val(k1) = fzero(F,k1);
end
Uh_val = uniquetol(h_val, 0.01)
There are infinity of solutions. These are some of them.
.
2 件のコメント
Star Strider
2022 年 5 月 7 日
The value of ‘F’ at that value of ‘h’ is not zero, so it will be necessary to adjust other parameters of the funciton in order to satisfy that requirement —
theta=30;
k=360;
h=0.5555;
F = @(h) ((cosd(theta).*(sind(k*h).^2)) - (2*(sind(k.*h.*cosd(theta)).^2)));
fval = F(0.555)
I leave that to you.
.
その他の回答 (0 件)
参考
製品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!