Unsure how to use initial conditions in characteristic value problem $u_{xx} - 7u_{tx} + 12u_{tt} = 0$
$u(t,x) = sinx$ on $t+3x=0$
$u(t,x) = x$ on $t+4x = 0$

I have factorized the above equation, which gives me the result:
$u(t,x) = f(t+3x) + g(t+4x)$
But, I don't know how to proceed. All I have is:
$f(0) + g(t+4x) = sinx$ and $f(t+3x) + g(0) = x$
If the conditions were $u(t,x)$ and $u_t(t,x)$, I'd be able to differentiate one to find $f$ and $g$. But here, I am completely lost. Any help would be appreciated!

$u_{xx} - 7u_{tx} + 12u_{tt} = 0$              ...$(i)$
$u(t,x) = sinx$ on $t+3x=0$ ...$(ii)$
$u(t,x)=x$ on $t+4x=0$ ...$(iii)$
First of all, we observe from equations $(ii)$  and $(iii)$   that $u(t, x)$ is independent of $t$ for all $t$ and $x$, hence the second and third term of equation $(i)$ are zero.
differentiating the equations $f(0) + g(t+4x) = sinx\implies\frac {\text d}{\text dx} g(t+4x)=\cos x$ ...$(iv)$
and $f(t+3x)+g(0)=x \implies \frac {\text d}{\text dx} f(t+3x)=1$ ...$(v)$
adding $(iv)$ and $(v)$ gives
$\frac {\text d}{\text dx} v(t,x)=\cos x+1$
$\implies u_{xx}=-\sin x$
