$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
Consider eigenvalues $\lambda=k^4$ of \begin{align} &X^{IV}-\lambda X=0, \label{eq-4.2.V6-1}\\ &X(0)=X'(0)=0,\label{eq-4.2.V6-2}\\ &X(l)=X'(l)=0.\label{eq-4.2.V6-3} \end{align} Then \begin{equation} X(x)= A\cosh(kx) + B\sinh (kx) +C\cos(kx) +D\sin(kx). \label{eq-4.2.V-4} \end{equation} then (\ref{eq-4.2.V6-2}) imply that $C=-A$, $D=-B$ and \begin{equation} X(x)= A(\cosh(kx) -\cos(x))+ B(\sinh (kx) -\sin(kx)). \label{eq-4.2.V-5} \end{equation} Plugging into (\ref{eq-4.2.V6-3}) we get \begin{align*} &A(\cosh(kl) -\cos(kl))+ B(\sinh (kl) -\sin(kl))=0,\\ &A(\sinh(kl)\ +\sin (kl))+ B(\cosh (kl) -\cos(kl))=0. \end{align*} Then determinant must be $0$: $$ (\cosh(kl) -\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 $$ which is equivalent to \begin{equation} \cosh(kl)\cdot\cos(kl)=1. \label{eq-4.2.V-6} \end{equation} We solve it graphically (see blue and red lines):
Consider eigenvalues $\lambda=k^4$ of \begin{align} &X^{IV}-\lambda X=0, \tag{1}\\ &X''(0)=X'''(0)=0,\label{eq-4.2.V6-7}\\ &X''(l)=X'''(l)=0.\label{eq-4.2.V6-8} \end{align} Then \begin{equation} X(x)= A\cosh(kx) + B\sinh (kx) +C\cos(kx) +D\sin(kx). \tag{4} \end{equation} then (\ref{eq-4.2.V6-7}) imply that $C=A$, $D=B$ and \begin{equation} X(x)= A(\cosh(kx) +\cos(x))+ B(\sinh (kx) +\sin(kx)). \label{eq-4.2.V-9} \end{equation} Plugging into (\ref{eq-4.2.V6-8}) we get \begin{align*} &A(\cosh(kl) +\cos(kl))+ B(\sinh (kl) +\sin(kl))=0,\\ &A(\sinh(kl)\ +\sin (kl))+ B(\cosh (kl) +\cos(kl))=0. \end{align*} Then determinant must be $0$: $$ (\cosh(kl) +\cos(kl))^2 -(\sinh^2 (kl) -\sin^2(kl))=0 $$ which is equivalent to \begin{equation} \cosh(kl)\cdot\cos(kl)=-1. \label{eq-4.2.V-10} \end{equation} We solve it graphically (see blue and green lines above).