Trigonometric non linear equation

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

Answer 1

[Star Strider]

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)

Uh_val = 1×79

1.0891 5.1140 9.8927 14.3144 18.4859 23.0537 27.0778 31.3009 35.8810 39.9061 45.7425 50.6869 54.7193 58.9369 64.0490 69.9192 73.9434 78.5763 82.6771 88.9498 92.9735 97.6948 101.7456 105.5893 110.3674 115.3914 119.5467 124.6211 128.8515 132.8778

There are infinity of solutions. These are some of them.

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[Sun Heat]

Thanks for replying..

will you please tell me the function which provide the value of 'h==0.555' as a answer

if we put same value as

theta=30; k=360;

((cosd(theta).*(sind(k*h).^2)) - (2*(sind(k.*h.*cosd(theta)).^2)))==0.0730

[Star Strider]

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)

fval = 0.0699

I leave that to you.

.