You expressed it correctly as a Heaviside function now you find the Laplacian of both sides.
The right hand side becomes,
[-y'(0)-sy(0)+s^2F(s)] + 4F(s) = WHATEVER.
Where, "WHATEVER" is the Laplacian of: 3sin(t) [1-u(t-pi)] -3sin(t)u(t-pi)
First you can write,
3sin(t) - 6sin(t)*u(t-pi)
Use trigonometric identity: sin(t) = -sin(t-pi)
6sin(t-pi)*u(t-pi) + 3sin(t)
Now the Laplace Transform of,
L{u(t-c)*f(t-c)} = exp(-cs) * F(s)
Thus,
L{6sin(t-pi)*u(t-pi)} = 6*exp(-pi*s)*L{sin(t)}
6*exp(-pi*s)*(1/(s^2+1^2))
So now, you have found equation in terms of Laplace transforms you just solve for F(s) an find the inverse Laplace transform.