= Solution
If $Q\in O(2)$, then $Q^TQ=I$, so
$$
(\det Q)^2=1
$$
and $\det Q=\pm1$. If $\det Q=1$, its first column is $(\cos\theta,\sin\theta)^T$ for some $\theta$, and orthonormality plus positive orientation force
$$
Q=
\begin{pmatrix}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta
\end{pmatrix}.
$$
Thus every element of $SO(2)$ is a rotation.
Let $J=\operatorname{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and
$$
\boxed{O(2)=SO(2)\sqcup SO(2)J}.
$$
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