Solving one equation with one unknown and get all possible solutions

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Hassan Alkomy
Hassan Alkomy 2018-10-25
评论： Hassan Alkomy 2019-1-8
I have an equation and I need to get its solution. I think it has more than one solution, but using the command (solve) I can get only one solution.
Actually, it is expected to get real and complex solutions, but I am interested on the real solutions only.
How can I get this solution in Matlab.
the required unknown is (alphap) and my equation and the command that I have used is:
m=15; Ki=1.3908e+06; B=0.945e-1; db=0.79e-2; alphao=.2618;
x = solve(Pr == m*Ki*(B*db*(cos(alphao)/cos(alphap)-1))^(3/2)*sin(alphap),alphap)
The answer is:
x = 0.37336926931567958392238007768557i

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采纳的回答

Stephan
Stephan 2019-1-4
编辑：Stephan 2019-1-4
Hi,
getting all possible soultions is a hard job, because you have an infinite bunch of real solutions:
You find them for example by using fzero:
m=15;
Ki=1.3908e+06;
B=0.945e-1;
db=0.79e-2;
alphao=.2618;
Pr = 10;
format long
fun = @(alphap)Pr - m*Ki*(B*db*(cos(alphao)/cos(alphap)-1))^(3/2)*sin(alphap)
x1 = fzero(fun,0.5)
This code results in:
x1 =
0.542034560066698
If you want more solutions just add or subtract integer multiples of 2*pi. Then you can construct as many real solutions as you want by yourself:
x2 = fzero(fun,2*pi+x1)
is_it_2_pi = (x2-x1)/(2*pi)
gives:
x2 =
6.825219867246284
is_it_2_pi =
1
Best regards
Stephan
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Walter Roberson
Walter Roberson 2019-1-8
The way your problem is constructed, with the sin(alphap) and cos(alphap) you would expect the results to repeat exactly every 2 π radians, since sin(alphap + 2*pi) = sin(alphap) and cos(alphap + 2*pi) = cos(alphap) . In theory. In practice due to round-off error it does not hurt to use the +2*pi as the starting point and use fzero() to confirm the exact location to within numeric bounds.

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更多回答（1 个）

Vineeth Nair
Vineeth Nair 2018-10-30
编辑：Vineeth Nair 2018-10-30
To get only real values use following command >>solve(equation, variable, 'Real', true)
You can read more about the solve function here .

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