General solution for $ x >-t $: $ u = \psi(x+2t) + \phi(x-2t) $
Given initial conditions, we have $ \psi(x) = \phi(x) = 0 $ so $$ u = 0, x > 2t $$
Using the boundary conditions for $ -t < x < 2t $, ie where $\phi(x), x<0 $
$$ \psi'(t) + \phi'(-3t) = \sin(t) $$
From previous solution $$ \psi(x) = 0 \implies \psi'(x) = 0 $$
So $$ \phi(t) = -\cos(\frac{-t}{3}) + constant, t <0 $$
$$ \phi(t) = -\cos(\frac{t}{3}) + constant, t<0 $$
$$ \phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant $$
For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:
$$ 0 = -\cos(0) + c \implies c = 1 $$
So $ u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t $
