Condition for $\Delta t$ so that the solution of the characteristic polynomial satisfies $|\epsilon|\leq 1 $

Question

Show that the characteristic polynomial is given by:

$\epsilon-1+4b_1$sin$^2(\frac{\Delta x}{2})=0$

and what is the condition for $\Delta{t}$ so that the roots of the characteristic polynomial satisfies $|\epsilon | \leq 1$

EDIT:

The difference equation considered is:

$P^{l+1}_{k}=wP^{l}_{k-1}+(1-2w)P^{l}_{k}+wP^{l}_{k+1}$ where $w= \frac{\Delta t}{constant*(\Delta x)^2}$

From here onwards the steps of the Neumann stability analysis is followed, which means the following form is substituted:

$P_{k,l}=\epsilon^lexp(-jk\Delta x)$. Now the question is complete.