= Solution
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In SI units the <Maxwell's equations> in vacuum, with sources, are
$$
\nabla\cdot E=\rho/\epsilon_0,\quad \nabla\cdot B=0,\quad
\nabla\times E=-\partial_tB,\quad
\nabla\times B=\mu_0j+\mu_0\epsilon_0\partial_tE.
$$
Taking the divergence of the last equation and using <Gauss's law> gives $\partial_t\rho+\nabla\cdot j=0$, the local <conservation of electric charge>. Thus sources that violate <charge conservation> cannot satisfy all the <Maxwell's equations>. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; <Faraday's law> similarly preserves $\nabla\cdot B=0$.
With $\rho=j=0$, use $\nabla\times(\nabla\times E)=\nabla(\nabla\cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-\mu_0\epsilon_0\partial_t^2E$. Applying the same calculation to $B$ yields
$$
\boxed{\Delta E-c^{-2}E_{tt}=0,\qquad
\Delta B-c^{-2}B_{tt}=0,\quad c=(\mu_0\epsilon_0)^{-1/2}.}
$$
A monochromatic <plane wave> has one <angular frequency> $\omega$ and constant-phase planes $\mathbf K\cdot r=\omega t-\text{constant}$, normal to the <wavevector> $\mathbf K$. Its <wavelength> is $2\pi/|\mathbf K|$.
For a nontrivial traveling wave with $\omega\neq0$, substitution of the displayed <electric field> into the divergence equation gives $\mathbf K\cdot e_1=\mathbf K\cdot e_2=0$. <Faraday's law> is satisfied by
$$
\boxed{B=\frac1\omega\,\mathbf K\times E,\qquad
\omega^2=c^2|\mathbf K|^2,\qquad \mathbf K\cdot e_1=\mathbf K\cdot e_2=0.}
$$
Indeed $\partial_tB=\mathbf K\times E_\phi$ and $\nabla\times E=-\mathbf K\times E_\phi$. The magnetic divergence is zero, while <Ampère's law> reduces to $|\mathbf K|^2E_\phi/\omega=(\omega/c^2)E_\phi$, giving the <dispersion relation>. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the <polarization>.
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