Trigonometric non linear equation
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Sun Heat
am 7 Mai 2022
Bearbeitet: Sun Heat
am 20 Mai 2022
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
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Star Strider
am 7 Mai 2022
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 Kommentare
Star Strider
am 7 Mai 2022
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.
.
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