We use the property of Laplace transform: $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$, where $F(s) = \mathcal{L}\{f(t)\}$.
In this case, $a=-2$ and $f(t) = \cos(4t)$. We know that $\mathcal{L}\{\cos(4t)\} = \frac{s}{s^2+16}$.
Therefore, $\mathcal{L}\{e^{-2t}\cos(4t)\} =
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