= Solution
This is a <plane electromagnetic wave> with <linear polarization>. $\mathbf E$ is the <electric field>; $E_0$ its peak <amplitude>; $x$ the position along propagation; $t$ the time; $k=2\pi/\lambda$ the <wavenumber>; and $\omega=2\pi f$ the <angular frequency>. Here $\lambda$ is the <wavelength>, $f$ the <frequency>, and $\hat{\mathbf y}$ the <unit vector> in the direction of the <electric field>. Constant <phase> $kx-\omega t$ moves along $+x$ with <phase velocity> $\omega/k$.
Put $\theta=kx-\omega t$. The differential <Faraday's law of induction> gives
$$
\nabla\times\mathbf E=-kE_0\sin\theta\,\hat{\mathbf z}=-\frac{\partial\mathbf B}{\partial t}.
$$
The accompanying oscillating <magnetic field>, excluding any independent static background, is therefore $\mathbf B=(k/\omega)E_0\cos\theta\,\hat{\mathbf z}$. To determine the propagation speed, also use the source-free <Ampere-Maxwell law>:
$$
\nabla\times\mathbf B=\mu_0\epsilon_0\frac{\partial\mathbf E}{\partial t},\qquad
k\frac{k}{\omega}E_0\sin\theta=\mu_0\epsilon_0\omega E_0\sin\theta.
$$
Thus
$$
\boxed{v=\frac\omega k=\frac1{\sqrt{\mu_0\epsilon_0}}=c,\qquad \mathbf B(x,t)=\frac{E_0}{c}\cos(kx-\omega t)\,\hat{\mathbf z}.}
$$
The <electric field> and <magnetic field> oscillate in <phase>, perpendicular to one another and to propagation; $\mathbf E\times\mathbf B$ points along $+x$.
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