solve command help needed.

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Divya Patel
Divya Patel 2021-3-13
评论： Walter Roberson 2021-3-13
here is my code,only 5 lines, and i m trying to solve for b,
syms b m theta
m = 5;
theta = 20;
eq = tand(theta)==(2*(1/tand(b)))*(((m*m*(sind(b)*sind(b))-1))/(m*m*(1.4+(cosd(b)*cosd(b)))+2)) ;
solve(eq,b)
the answer i get is this ==> -(log(root(z^6 - z^4*(1238888939898706904063102115185851/2095413426666905974699416890572625 + 2864291029470578218930659559211008i/2095413426666905974699416890572625) - z^2*(2168555873514773727160036100864837/2095413426666905974699416890572625 - 2244186785976947882873506458763264i/2095413426666905974699416890572625) + (78442739762537124403601334665223/83816537066676238987976675622905 + 29528773499696682669388242878464i/83816537066676238987976675622905), z, 1))*180i)/pi
now idk where z comes from since its not part of my code, but i also saw some imaginary parts so since i wanted a real number i tried this
syms b m theta
m = 5;
theta = 20;
eq = tand(theta)==(2*(1/tand(b)))*(((m*m*(sind(b)*sind(b))-1))/(m*m*(1.4+(cosd(b)*cosd(b)))+2)) ;
solve(eq,b,'real',true)
here i get the answer
ans = (180*(2*pi*k + atan2(x, 1) - atan2(-x, 1)))/pi
now its a smaller eq but idk what the k or x values to be,and where they come from
the ideal answer should be 29.8 for b.
so any one know whats happening and how to solve this.

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采纳的回答

Walter Roberson
Walter Roberson 2021-3-13
Ran in:
More solutions than you can shake a stick at.
syms b m theta
m = 5;
theta = 20;
eqn = tand(theta)==(2*(1/tand(b)))*(((m*m*(sind(b)*sind(b))-1))/(m*m*(1.4+(cosd(b)*cosd(b)))+2)) ;
sol = solve(eqn, b, 'returnconditions', true)
sol = struct with fields:
b: [1×1 sym] parameters: [1×2 sym] conditions: [1×1 sym]
sol.b
ans = 
sol.parameters
ans = 
sol.conditions
ans = 
k = 0;
z1 = solve(subs(sol.conditions, sol.parameters(1), k));
sol6_0 = simplify(subs(subs(sol.b, sol.parameters(1), k), sol.parameters(2), z1))
sol6_0 = 
vpa(sol6_0)
ans = 
k = 1;
z1 = solve(subs(sol.conditions, sol.parameters(1), k));
sol6_1 = simplify(subs(subs(sol.b, sol.parameters(1), k), sol.parameters(2), z1))
sol6_1 = 
vpa(sol6_1)
ans = 
k = -1;
z1 = solve(subs(sol.conditions, sol.parameters(1), k));
sol6_m1 = simplify(subs(subs(sol.b, sol.parameters(1), k), sol.parameters(2), z1))
sol6_m1 = 
vpa(sol6_m1)
ans = 
  2 个评论
Divya Patel
Divya Patel 2021-3-13
hello sir the "vpa(sol6_0)"
does give the right answer, well out of the 6 values it outputs one of the value is the usefull needed one, but if possible can you explain how the code exicuted to get me this, as i want to stich this value into a bigger recurring code.
Thank You
Walter Roberson
Walter Roberson 2021-3-13
Ran in:
You probably don't really care. What you probably want is more like
syms m theta
syms b
assume(b >= -90 & b <= 90)
m = 5;
theta = 20;
eq = tand(theta)==(2*(1/tand(b)))*(((m*m*(sind(b)*sind(b))-1))/(m*m*(1.4+(cosd(b)*cosd(b)))+2)) ;
vpasolve(eq, b)
ans = 
33.175834649454488192870542271753
all of the values in sol6_0, sol6_1, sol6_m1, and many others, are valid solutions for your equation, and you should be using them all -- unless you have physical reason to constrain the angles.

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