$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ $\newcommand{\bR}{\mathbb{R}}$
Fourier series were introduced to solve problems allowing separation of variables. However they can be used without it.
Example 1. Consider problem \begin{align} &u_{tt}-2au_{tx}-u_{xx}=0&&0<x<\pi, \label{eq-4.D-1}\\ &u|_{x=0}=u|_{x=\pi}=0 && \label{eq-4.D-2}\\ &u_{t=0}+g(x),\ u_{t}|_{t=o}=h(x). \label{eq-4.D-3} \end{align} Actually it will be hyperbolic as long as coefficient at $u_{xx}$ belongs to $(-\infty, a^2)$ but we take it $-1$.
Skip (\ref{eq-4.D-3}). Obviously, separation of variables does not work. However, let us try $u(x,t)= e^{ikt}X(x)$ with $k\in\bC$. Then we get \begin{gather} -k^2 X -2aik X' -X''=0, \label{eq-4.D-4}\\ X(0)=X(\pi)=0. \label{eq-4.D-5} \end{gather} This is an eigenvalue/eigenfunction problem for pencils, which is rather arcane. However if we introduce $\boldsymbol{X}=\left(\begin{matrix}v\\ w\end{matrix}\right)$ with $v=X'$, $w=ikX$ then we can rewrite it as \begin{align} &k\boldsymbol{X}=k\begin{pmatrix} v \\ w \end{pmatrix} = \begin{pmatrix} kX' \\ ik^2 X\end{pmatrix} = \begin{pmatrix} kX' \\ 2akX' -X''\end{pmatrix}= A i\partial_x \boldsymbol{X}, \label{eq-4.D-6}\\[7pt] &w(0)=w(\pi)=0\label{eq-4.D-7} \end{align} with \begin{gather} A= \begin{pmatrix} 0 & 1 \\ 1 & 2a \end{pmatrix}. \label{eq-4.D-8} \end{gather} One can see easily $A$ is a Hermitean matrix and that operator $Ai\partial_x $ with these boundary condition is symmetric operator in $L^2([0,\pi])$. One can prove also that it is self-adjoint operator and one can also prove that it has a discrete spectrum $\subset \mathbb{R}$. So we got a complete system of orthogonal eigenfunctions.
This spectrum has multiplicity $1$. Solving ODE (\ref{eq-4.D-4}) we get that $X= (A\cos (r x) + B\sin (rx)) e^{-a ik x}$ with $r=\sqrt{a^2+k^2}$ and to satisfy (\ref{eq-4.D-7}) we need $r=n$ with $n=1,2,\ldots$: \begin{gather} X_n= \sin (nx) e^{-a i k_n x}\quad n=1,2,\ldots,\label{eq-4.D-9}\\ k_n= -a \pm \sqrt{n^2-a^2} ;\label{eq-4.D-10} \end{gather} recall that $a^2<1$.
Then \begin{gather*} u_n= e^{ik_n (t-ax)}\sin (nx)= e^{-ai (t-ax)} e^{\pm i (t-ax) \sqrt{n^2-a^2}} \sin(nx) \end{gather*} and repacing the second factor via sines and cosines and taking sum over $n$ with arbitrary coefficients we arrive finally to \begin{gather} u(x,t) =\sum _{n=1}^\infty \Bigl(A_n \cos((t-ax)\sqrt{n^2-a^2}) + B_n \sin((t-ax)\sqrt{n^2-a^2}) \Bigr) e^{-ai (t-ax)} \sin(nx) . \label{eq-4.D-11} \end{gather} These coefficients $A_n,B_n$ could be found from initial conditions (\ref{eq-4.D-3}).
Example 2. Consider \begin{align} &\Delta^2 u =0 &&0<x<\pi, \ 0<y < b,\label{eq-4.D-12}\\ &u|_{x=0}=u_{xx}|_{x=0}= u|_{x=\pi}=u_{xx}|_{x=\pi}=0,\label{eq-4.D-13}\\ & B_1u|_{y=0}=g, \quad B_2u|_{y=b}=h\label{eq-4.D-14} \end{align} where $B_j$ are some boundary operators; recall that on each end $y=0$ and $y=b$ we need to have two boundary conditions, so those are $2\times 1$ matrix operators, f.e. $B_1= \begin{pmatrix} 1 \\ \partial_y \end{pmatrix}$ and $B_2=\begin{pmatrix} 1 \\ \partial^2_y \end{pmatrix}$.
Skipping (\ref{eq-4.D-14}) and plugging into (\ref{eq-4.D-12}) $u(x,y)=X(x)Y(y)$ leads us to \begin{gather} X^{IV}Y+ 2X''Y'' +XY^{IV}=0 \iff \frac{X^{IV}}{X} + 2\frac{X''}{X}\cdot \frac{Y''}{Y}+\frac{Y^{IV}}{Y}=0 \label{eq-4.D-15} \end{gather} with the middle term preventing separation. However, taking $X=\sin(nx)$ with $n=1,2,\ldots$ we satisfy (\ref{eq-4.D-13}) and transform (\ref{eq-4.D-15}) into \begin{gather} Y^{IV}-2n^2 Y'' + n^4Y = k^4Y \end{gather} and then \begin{gather} Y(y)= A_n cos(ny) + B_n \sin (ny) + C_n\cosh (ny) +D_n\sinh(ny) \end{gather} and we get \begin{gather} u(x,y)=\sum_{n=1}^\infty \Bigl(A_n cos(ny) + B_n \sin (ny) + C_n\cosh (ny) +D_n\sinh(ny)\Bigr)\sin (nx) \end{gather} and we recover coefficients $A_n,B_n,C_b,D_n$ from four boundary conditions (\ref{eq-4.D-14}).