solve command help needed.

2021 3 13
1 个回答
回答已采纳
更新时间：2021 3 13
5 次查看（30 天）

0 个投票

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.

请先登录，再回答此问题。

 采纳的回答

Walter Roberson
Walter Roberson 2021-3-13
Ran in:

1 个投票

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
显示 无 隐藏 无

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:
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.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Assumptions 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by