For which parameters are the solution stable for an ODE? We've got
$$x''+bx'+k^2\sin x=0$$
for which $b,k$ are real and $k$ is not equal to $0$.
For which $b$ and $k$ do we got a stable, asymptotically stable and unstable solution?
 A: First, write this as a first order vector equation by letting $y = x'$, giving the equation
$$ \mathrm x' =  \begin{pmatrix} x \\ y \end{pmatrix}' = \begin{pmatrix} y \\ -k^2\sin x - y \end{pmatrix} $$
Obviously, the presence of $\sin x$ indicates that this is a nonlinear equation, so we'll need to linearize about our equilibrium points to get any information about them. To do that, we need to find the equilibria, so we set the right hand side to $0$.
$$ y^* = 0 $$
$$ x^* = n\pi $$
Then, we use the Jacobian as the coefficient matrix to get a linear equation which approximates our nonlinear near the equilibria.
$$  \mathrm x' = \mathrm J \mathrm x $$
$$ \mathrm J = \begin{pmatrix} 0 & 1 \\ -k^2\cos n\pi & -b \end{pmatrix} $$
Because $\cos n\pi$ oscillates between $1$ and $-1$, let's split it up into two cases: $n_0 = 2k$ and $n_1=2k+1$. Correspondingly, we have $x^*_0 = 2k\pi$ and $x^*_1 = (2k+1)\pi$, and the Jacobian matrices
$$ \mathrm J_0 = \begin{pmatrix} 0 & 1 \\ -k^2 & -b \end{pmatrix} $$
and
$$ \mathrm J_1 = \begin{pmatrix} 0 & 1 \\ k^2 & -b \end{pmatrix} $$
Now, the only information we really need to speak of stability for this system is the sign of the real part of the eigenvalues, so let's get those eigenvalues.

$ \mathrm J_0 $ has eigenvalues $\lambda_{1,2}$ that are roots of
$$ \lambda^2 + b\lambda + k^2 $$
which come in the form
$$ \lambda_{1,2} = \frac{-b\pm\sqrt{b^2-4k^2}}{2} $$
If $b^2 \leq 4k^2$, the square root term doesn't contribute to the real part, so the signs will both be the opposite of $b$. If $b^2 > 4k^2$, then we must have that $|b| \geq \sqrt{b^2 - 4k^2}$, so  again the signs are both the opposite of $b$. This means that when $b > 0$, these equilibria will be stable for all $k$, and for $b < 0$ they will be unstable for all $k$. When $b = 0$, we have a pair of pure imaginary roots, so we'll have concentric periodic orbits around the equilibrium, at least locally. Unless also $k = 0$, then we have unstable parallel line orbits (this is $x'' = 0$).

$ \mathrm J_1 $ has eigenvalues $\lambda_{1,2}$ that are roots of
$$ \lambda^2 + b\lambda - k^2 $$
which come in the form
$$ \lambda_{1,2} = \frac{-b\pm\sqrt{b^2+4k^2}}{2} $$
These roots are always real, and since $\sqrt{b^2+4k^2} \geq |b|$, the two roots will have opposite sign. This will give us an unstable saddle when $b$ and $k$ are not both $0$, and when they are it will be the same as above.

So, to answer the question, there are always unstable solutions, and there are only stable solutions when $b$ is positive, regardless of $k$.
