How to solve single variable nonlinear trigonometric function?

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Jintu K James
Jintu K James 2021-9-6
评论： Jintu K James 2021-9-7
I am trying to solve a non linear equation as given below.
I am expecting β2 to be between 35 and 50 degrees. All angles are in degrees.
I tried using and using functions. But the reults were not in that range . How to solve this issure.
Thanks in advance.

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采纳的回答

Abolfazl Chaman Motlagh
Ran in:
maybe you forgot to use "d" in trigonometric functions.
quick answer without optimizing the code:
m = 0.7:0.01:1 ;
b1 = 34.65 ;
theta = 16;
syms b;
f =@(b,m) m - (1-(tand(b-b1).*sind(theta))./(sind(b1).*sind(b-theta))).^2;
for i=1:numel(m)
B(i) = vpasolve(f(b,m(i)) == 0, b);
end
min(B)
ans = 
34.65
max(B)
ans = 
43.489918649470219406073636845804
plot(B)
  2 个评论
Abolfazl Chaman Motlagh
Abolfazl Chaman Motlagh 2021-9-6
Ran in:
or use fsolve with initial guess:
m = 0.7:0.01:1 ;
b1 = 34.65;
theta = 16;
f =@(b) m - (1-(tand(b-b1).*sind(theta))./(sind(b1).*sind(b-theta))).^2;
B = fsolve(f , 40*ones(1,31));
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
min(B)
ans = 34.6500
max(B)
ans = 43.4899
plot(B)
Jintu K James
Jintu K James 2021-9-7
Thank you very much for the help. It worked.
I used 'd' in trignometric functions, but i forgot to put 'm' in the f =@(b,m) command while using vpasolve.

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