Let $x \in [0,2\pi)$.
Define $A_x = \left (\begin{matrix} \cos(x) & -\sin(x) & 0&0 \\ \sin(x) & \cos (x) & 0 &0 \\ 0 & 0 & \cos(x) & -\sin(x) \\ 0& 0& \sin(x) & \cos(x) \end{matrix} \right)$.
Then $\forall x\in [0,2\pi), A_x$ is well defined.
My question is, does there exist some $x\in [0,2\pi)$ such that $A_x$ is not a rotation? Here is my definition of rotation:
An orthogonal operator $T$ on a nonzero real inner product space $V$ is a rotation if
- $T$ is the identity map, or
- there exists a 2-dimensional subspace $W$ of $V$, an orthonormal basis $\{x_1,x_2\}$ for $W$ and some $\theta \in [0,2\pi)$ such that \begin{cases} T(x_1) = \cos(\theta)x_1+ \sin(\theta)x_2,\\ T(x_2) = -\sin(\theta)x_1 + \cos(\theta)x_2,\\ T(y)=y\ \text{ for all }\ y\in W^{\perp}. \end{cases}
One way to detect if the transformation is leaving the orthogonal complement alone is to check the eigenvalues. If it really is leaving a two dimensional subspace invariant, it should have eigenvalue 1 with at least multiplicity 2.
A computation shows that the eigenvalues are $\cos(x)\pm i\sin(x)$, each of them having multiplicity two. So, the only way there can be an invariant subspace of dimension at least 2 is for $x=0$, when you get the identity.
I guess, as you hinted in the comments above, that this matrix results as a composition of $$\begin{bmatrix}\cos(x)&-\sin(x)&0&0\\\sin(x)&\cos(x)&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&\cos(x)&-\sin(x)\\0&0&\sin(x)&\cos(x)\end{bmatrix}$$
You can easily see that these matrices individually have two-dimensional eigenspaces for the eigenvalue $1$. The other part of the requirement for rotations about two orthonormal vectors is fulfilled by the columns containing sines and cosines.