Are these two PDEs equivalent somehow?

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I have the one dimensional wave equation for $f(x,y)$ $(1)$: $$\frac{1}{c^2} \frac{\partial^2 f}{\partial t^2} = \frac{\partial^2 f}{\partial x^2}$$ and a system of PDE for $f(x,y)$ and $g(x,y)$ $(2)$:

\begin{equation} \frac{1}{c} \frac{\partial f}{\partial t} = \frac{\partial g}{\partial x} \\ \frac{1}{c} \frac{\partial g}{\partial t} = \frac{\partial f}{\partial x} \end{equation}

I know that any solution to $(2)$ for $f$ is a solution to $(1)$.

My question is: Is any solution to $(1)$ for $f$ also a solution to $(2)$ for $f$?

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At least locally, yes: If $f$ satisfies (1) in an open ball, then

$$\frac{\partial}{\partial t} \left( \frac{1}{c} \frac{\partial f}{\partial t}\right) = \frac{\partial}{\partial x} \left( c \frac{\partial f}{\partial t}\right)$$

Since the open ball is simply connected, there is a function $g$ so that

\begin{align} \frac{\partial g}{\partial x} &= \frac 1c \frac{\partial f}{\partial t},\\ \frac{\partial g}{\partial t} &= c \frac{\partial f}{\partial x} \end{align}

and this is just (2).