= Solution
Take the curl of the vacuum <Faraday law> and use the vacuum <Ampere-Maxwell law>:
$$
\nabla\times(\nabla\times\mathbf E)=-\partial_t(\nabla\times\mathbf B)=-\mu_0\varepsilon_0\partial_t^2\mathbf E.
$$
Since the electric divergence vanishes, the double-curl identity makes the left side $-\Delta\mathbf E$. Therefore the <electromagnetic wave equation> is
$$
\boxed{\Delta\mathbf E-\frac1{c^2}\partial_t^2\mathbf E=0,\qquad c=\frac1{\sqrt{\mu_0\varepsilon_0}}.}
$$
The same calculation interchanging the fields gives the wave equation for $\mathbf B$.
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