Using Maxwell's equations to find $\mathbf{B}$ The $\mathbf{E}$ component of an electromagnetic wave in free space is:
$$\mathbf{E}(x, t) = E_0 \cos{(kx-\omega t)} \hat{\bf y}$$
How do I find the corresponding $\mathbf{B}$ component using one of Maxwell's equations (in differential form)?
I assume that I must use one of either
$$\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$
or
$$\nabla\times\mathbf{B} = \epsilon_0\mu_0\frac{\partial\mathbf{E}}{\partial t}$$
(probably the second one?)
I also know that I am looking for an expression in the form of $$\mathbf{B}(x, t) = B_0 \cos{(kx-\omega t)} \hat{\bf z},$$
but I am not sure about the intermediate steps.
 A: 
I assume that I must use one of either ... or ...
(probably the second one?)

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.
A: In a region with no charges ($\rho = 0$) and no currents ($\mathbf J=\mathbf 0$) such as in a vacuum, Maxwell's equations reduce to:
$$\begin{align}
  \boldsymbol \nabla \cdot \mathbf{E} &= 0 \quad & \boldsymbol\nabla \times \mathbf{E} &= -\frac{\partial\mathbf B}{\partial t}, \\
  \boldsymbol\nabla \cdot \mathbf{B} &= 0 \quad & \boldsymbol\nabla \times \mathbf{B} &= \mu_0\varepsilon_0 \frac{\partial\mathbf E}{\partial t}.
\end{align}.$$
I suggest to see this link:
https://mathworld.wolfram.com/dAlembertsSolution.html
Taking the curl of the curl equations, you will get
$$\begin{align}
  \mu_0\varepsilon_0  \frac{\partial^2 \mathbf{E}}{\partial t^2} - \nabla^2 \mathbf{E} = 0 \\
  \mu_0\varepsilon_0 \frac{\partial^2 \mathbf{B}}{\partial t^2} - \nabla^2 \mathbf{B} = 0
\end{align}
$$
The quantity $\mu_0\varepsilon_0$ has the dimension of (time/length)$^2$. Defining
$c = (\mu_0 \varepsilon_0)^{-1/2}$, the equations above have the form of the standard wave equation's
$$\begin{align}
  \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - \nabla^2 \mathbf{E} = 0 \\
  \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - \nabla^2 \mathbf{B} = 0
\end{align}$$
I remember that the d'Alembert operator is:
$$\square = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 .$$
A: You can integrate the first equation in time to obtain:
\begin{align}
\int \nabla\times\mathbf{E} dt =  \mathbf{B}(t) -   \mathbf{B}(0)
\end{align}
If if you have already an expression for  $\mathbf{E}(t)$, you can just substitute it into the above equation and you'll obtain an expression for  $\mathbf{B}(t)$.
