How to solve single variable nonlinear trigonometric function?

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.

Answer 1

[Abolfazl Chaman Motlagh]

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)

[Abolfazl Chaman Motlagh]

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]

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.