= Solution
In vacuum the source-free equations are
$$
\nabla\cdot\mathbf E=\nabla\cdot\mathbf B=0,\qquad\nabla\times\mathbf E=-\mathbf B_t,\qquad\nabla\times\mathbf B=c^{-2}\mathbf E_t.
$$
Adding the divergence equations with weights $1,ic$ gives the complex divergence equation. Adding the curl equations in the same way gives $\nabla\times\mathbf W=(i/c)\mathbf W_t$. Conversely real and imaginary parts of these two complex equations recover all four real equations.
For the cosine ansatz, differentiation yields divergence $(\boldsymbol\Omega\cdot\mathbf K)\sin(\omega t-\mathbf K\cdot\mathbf x)$ and curl $-(\boldsymbol\Omega\times\mathbf K)\sin(\omega t-\mathbf K\cdot\mathbf x)$. The time <derivative> is $-\omega\boldsymbol\Omega\sin(\omega t-\mathbf K\cdot\mathbf x)$. Since that sine is not identically zero, the required conditions are precisely
$$
\boxed{\boldsymbol\Omega\cdot\mathbf K=0,\qquad\boldsymbol\Omega\times\mathbf K=(i\omega/c)\boldsymbol\Omega.}
$$
Dotting the second equation with $\boldsymbol\Omega$ gives $\boldsymbol\Omega\cdot\boldsymbol\Omega=0$ because $\omega\ne0$. This dot product is complex bilinear, not Hermitian. Taking the <cross product> again with $\mathbf K$ and using transversality gives $-|\mathbf K|^2\boldsymbol\Omega=-\omega^2\boldsymbol\Omega/c^2$. Nonzero amplitude therefore requires $\boxed{|\mathbf K|^2=\omega^2/c^2}$.
Taking real and imaginary parts yields the full conditions
$$
\boxed{\mathbf e\cdot\mathbf K=\mathbf b\cdot\mathbf K=0,\quad\mathbf K\times\mathbf e=\omega\mathbf b,\quad\mathbf b\times\mathbf K=(\omega/c^2)\mathbf e.}
$$
They imply $\mathbf e\cdot\mathbf b=0$ and $|\mathbf e|=c|\mathbf b|$, together with the dispersion relation. Conversely a nonzero transverse $\mathbf e$, this dispersion relation and $\mathbf b=\mathbf K\times\mathbf e/\omega$ satisfy all the conditions.
The physical <fields> are $\mathbf E=\mathbf e\cos(\omega t-\mathbf K\cdot\mathbf x)$ and $\mathbf B=\mathbf b\cos(\omega t-\mathbf K\cdot\mathbf x)$. Each oscillates along one fixed real direction, so \b[both <fields> are linearly polarized]. Complex encoding of the two <fields> here does not mean either real <field> is circularly polarized.
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