Although this is a standard derivation, you frequently don't see it in introductory electromagnetism courses, maybe because those courses shy away from the heavy use of vector calculus. Here's the usual approach. We'll find a wave equation from Maxwell's equations.
Start with
$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$.
Take a partial derivative of both sides with respect to time. The curl operator has no partial with respect to time, so this becomes
$\nabla \times \frac{\partial\vec{E}}{\partial t} = -\frac{\partial^2\vec{B}}{\partial t^2}$.
There's another of Maxwell's equations that tells us about $\partial\vec{E}/\partial t$.
$\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$
Solve this for $\partial\vec{E}/\partial t$ and plug into the previous expression to get
$\nabla \times \frac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\frac{\partial^2 \vec{B}}{\partial t^2}$
the curl of curl identity lets us rewrite this as
$\frac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) - \nabla^2\vec{B}\right) = -\frac{\partial^2 \vec{B}}{\partial t^2}$
But the divergence of the magnetic field is zero, so kill that term, and rearrange to
$\frac{-1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \frac{\partial^2 \vec{B}}{\partial t^2} = 0$
This is the wave equation we're seeking. One solution is
$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$.
This represents a plane wave traveling in the direction of the vector $\vec{k}$ with frequency $\omega$ and phase velocity $v = \omega/|\vec{k}|$. In order to be a solution, this equation needs to have
$\frac{\omega^2}{k^2} = \frac{1}{\mu_0\epsilon_0}$.
Or, setting $v = 1/\sqrt{\mu_0\epsilon_0}$
$\frac{\omega}{k} = v$
This is called the dispersion relation. The speed that electromagnetic signals travel is given by the group velocity
$\frac{d\omega}{d k} = v$
So electromagnetic signals in a vacuum travel at speed $c = 1/\sqrt{\mu_0\epsilon_0}$.
Edit
You can follow the same steps to derive the wave equation for $\vec{E}$, but you will have to assume you're in free space, i.e. $\rho = 0$.
Edit
The curl of the curl identity was wrong, there's a negative number in there
In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.
But this is rather illusory. In relativity, the equations look quite different:
$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$
where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).
The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.
What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit more. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.
Best Answer
You need to use both of them.
From your expressions for $\mathbf{E}(x, t) $ and $\mathbf{B}(x, t)$ calculate $\nabla\times\mathbf{E}$ and $\frac{\partial \mathbf{B}}{\partial t}$. Insert the results into the first of your Maxwell equations. You will get $$E_0k=B_0\omega. \tag{1}$$
Likewise calculate $\nabla\times\mathbf{B}$ and $\frac{\partial\mathbf{E}}{\partial t}$. Insert the results into the second of your Maxwell equations. You will get $$B_0 k=\epsilon_0\mu_0 E_0\omega. \tag{2}$$
From (1) and (2) and a little bit of algebra you find $$B_0=E_0\sqrt{\epsilon_0\mu_0} \tag{3}$$ and $$k=\omega\sqrt{\epsilon_0\mu_0}. \tag{4}$$
(3) is the magnetic field amplitude you were looking for. And equation (4) tells you the speed of your electromagnetic wave is $\frac{1}{\sqrt{\epsilon_0\mu_0}}$ which happens to be equal to the speed of light $c=3\cdot 10^8$ m/s.