Laplace Transform with sin and cos

Question

Hi I am having trouble figuring out the solution of this Laplace transform:

$$L_t{(u(t- \pi)(2\cos(t)-3\sin(3t))}$$

Where I am stuck if I am even on the right track is:

$$L_t{(u(t- \pi)(2\cos(t)-3\sin(3t))}$$

$$=e^{- \pi (t)}L_t({2\cos((t)(t+ \pi)) -3\sin((3t)(t+ \pi)))}$$

and I am not sure where to go from here, any help would be appreciated. Thanks.

Answer 1

$\textbf{Hint:}$

You can start by using $L\big(u(t-\pi)(2\cos t-3\sin 3t)\big)=e^{-\pi s}L\big(2\cos(t+\pi)-3\sin(3(t+\pi)\big)=e^{-\pi s}L\big(-2\cos t+3\sin3t\big)$