Cannot understand Sturm-Liouville problem answer I have one Sturm-Louville question with answer, however I cannot understand the answer of this problem since it seem some part of the answer already simplified.
Below is the mentioned question and answer:
Question & Answer
Answer2
I dont understand this answer until the non-trivial solution about how it can end-up with 
A(1-cos(kpi))- Bsin(kpi) = 0
 A: They are considering the Sturm-Liouville problem
$$
                   y''+\lambda y = 0, \;\; y(0)=y(\pi),\;\;y'(0)=y'(\pi).
$$
Sturm-Liouville problems have non-trivial solutions only when the eigenvalues $\lambda$ are real. So they have broken the problem into cases where $\lambda < 0$, $\lambda > 0$, $\lambda=0$, which reduces the problem to $\lambda=-\mu^2$ where $\mu > 0$, and $\lambda=\mu^2$ where $\mu > 0$, and $\lambda=0.
The case under consideration in the posted question is where $\lambda=-\mu^2$ for $\mu > 0$. In this case, the ODE has solution
$$
                y(x)=Ae^{-\mu x}+Be^{\mu x}
$$
where $A$, $B$ are constants. The $\mu$ for which the above is an actual solution of the full problem are those $\mu$ for which
$$
      0 = y(\pi)-y(0) = (Ae^{-\mu\pi}+Be^{\mu\pi})-(A+B) \\
      0 = y'(\pi)-y'(0) = (-A\mu e^{-\mu\pi}+B\mu e^{\mu\pi})-(-A\mu+B\mu).
$$
Assuming $\mu \ne 0$ leads to a system of equations for the constants $A$, $B$:
$$
         (e^{-\mu\pi}-1)A+(e^{\mu\pi}-1)B = 0 \\
         (e^{-\mu\pi}-1) A-(e^{\mu\pi}-1) B = 0
$$
There is a non-trivial solution $A$, $B$ iff the determinant of the coefficient matrix for $A$, $B$ is $0$, i.e.,
$$
        -(e^{-\mu\pi}-1)(e^{\mu\pi}-1)-(e^{-\mu\pi}-1)(e^{\mu\pi}-1)=0 \\
      -2(e^{-\mu\pi}-1)(e^{\mu\pi}-1) = 0.
$$
There is no $\mu > 0$ for which the above is $0$ because $e^{\pm\mu\pi}\ne 1$. So there are no negative eigenvalues $\lambda=-\mu^2$ where $\mu > 0$.
