Need help solving equation in terms of a variable

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Kevin Bodell
Kevin Bodell 2021-4-20
评论： Walter Roberson 2021-4-21
I think this should be a simple solution, but I have an equation set equal to zero with two variables, "L" and "k." I'd like to solve the equation for "k" or ideally for "(k*L)/2." But am having trouble using the linsolve function. Below is the equation as well as the two linsolve approaches I've tried, both of which return a division by zero error.
eq8 = cos((L*k)/2)*cos((2^(1/2)*L*k)/4) - (2^(1/2)*sin((L*k)/2)*sin((2^(1/2)*L*k)/4))/2 == 0
%linsolve(eq8, k)
%linsolve(eq8, (k*L)/2)

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回答（1 个）

Walter Roberson
Walter Roberson 2021-4-21
Ran in:
syms L k Lk2
eq8 = cos((L*k)/2)*cos((2^(1/2)*L*k)/4) - (2^(1/2)*sin((L*k)/2)*sin((2^(1/2)*L*k)/4))/2 == 0
eq8 = 
K = solve(k*L/2 == Lk2, k)
K = 
eq8kL = subs(eq8, k, K)
eq8kL = 
vpasolve(eq8kL)
ans = 
fplot(lhs(eq8kL)-rhs(eq8kL), [-25 25])
Not a linear system; it is a periodic or quasi-periodic system, with an infinite number of solutions.
char(eq8kL)
ans = 'cos((2^(1/2)*Lk2)/2)*cos(Lk2) - (2^(1/2)*sin((2^(1/2)*Lk2)/2)*sin(Lk2))/2 == 0'
  1 个评论
Walter Roberson
Walter Roberson 2021-4-21
Maple says that the solutions are the θ such that
which looks reasonable from eq8kL .
Unfortunately that does not help find explicit formulas.

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