2. The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{0}=1, x_{1}=6, \\
x_{2}=x_{1}+\sin x_{1}=5.72 \cdots, \\
x_{3}=x_{2}+\cos x_{2}=6.56 \cdots .
\end{array}
$$

In general,
if $x_{n} \geqslant x_{n-1}$, then $x_{n+1}=x_{n}+\sin x_{n}$;
if $x_{n}<x_{n-1}$, then $x_{n+1}=x_{n}+\cos x_{n}$. For any $n \in \mathbf{Z}_{+}$, find the smallest constant $c$ such that $x_{n}<c$.
2. $c_{\min }=3 \pi$.

Firstly, note: if $x\frac{5 \pi}{2}$.
Thus, $\cos x_{k-1}2 \pi$, it is easy to prove by mathematical induction that $x_{n}>2 \pi,\left\{x_{n}\right\}$ is monotonically increasing.
Therefore, $x_{n+1}=x_{n}+\sin x_{n}(n \geqslant 3)$.
If $x_{n}<c<3 \pi\left(n \in \mathbf{Z}_{+}\right)$, then
$x_{n+1}-x_{n}=\sin x_{n}$
$\geqslant \min \left\{\sin x_{3}, \sin c\right\}(n \geqslant 3)$.
This indicates that $\left\{x_{n}\right\}$ is unbounded. Contradiction.
Thus, $c_{\min }=3 \pi$.