Problem 03 | First Shifting Property of Laplace Transform

$\mathcal{L}(\cos bt}) = \dfrac{s}{s^2 + b^2}$

$\mathcal{L}(\cos t}) = \dfrac{s}{s^2 + 1^2}$

$\mathcal{L}(\cos t}) = \dfrac{s}{s^2 + 1}$
 

Thus,
$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{(s + 3)^2 + 1}$

$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{(s^2 + 6s + 9) + 1}$

$\mathcal{L}(e^{-3t} \cos t}) = \dfrac{s + 3}{s^2 + 6s + 10}$           answer