Consider the following wave equation: $$\begin{align} u_{tt}-u_{xx} &=0, \quad 0<t<tx, k>1\\ u|_{t=0}&=\phi_0(x),\quad x\ge 0\\ u_t|_{t=0}&=\phi_1(x),\quad x\ge 0\\ u|_{t=kx}&=\psi(x)\end{align} $$ In which $\phi_0(0)=\psi(0)$.
The problem is that on part of the sector where $t>x$ d'Alembert's formula isn't applicable. It seems we will have to do some sort of extension or reflection, but how to start?
Given a point $(x,t)$ in the region $x<t<k\,x$, construct the parallelogram formed by the characteristic curves $x\pm t=1$ with vertices vertices in $(x,t)$, the line $t=k\,x$ and the line $t=0$ (the fourth vertex will fall in the region $0<x<t$). You know the value of $u$ at the last three of the vertices, and can find $u(x,t)$ using the fact that the sums of the values of $u$ at opposing vertices are equal (this is a property of solutions of the wave equation.)