2. Given the function
$$
f(x)=a \sin x+b \cos x \quad(a, b \in \mathbf{Z}),
$$

and it satisfies
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} .
$$

Then the maximum value of $a$ is . $\qquad$
2.3.
Let $A=\{x \mid f(x)=0\}, B=\{x \mid f(f(x))=0\}$.
Obviously, set $A$ is non-empty.
Take $x_{0} \in A$, i.e., $x_{0} \in B$, hence
$$
b=f(0)=f\left(f\left(x_{0}\right)\right)=0 \text {. }
$$

Thus, $f(x)=a \sin x(a \in \mathbf{Z})$.
When $a=0$, obviously, $A=B$.
Now assume $a \neq 0$, in this case,
$$
\begin{array}{l}
A=\{x \mid a \sin x=0\}, \\
B=\{x \mid a \sin (a \sin x)=0\} \\
=\{x \mid a \sin x=k \pi, k \in \mathbf{Z}\} .
\end{array}
$$

It is easy to see that $A=B$ if and only if for any $x \in \mathbf{R}$, $a \sin x \neq k \pi(k \in \mathbf{Z}, k \neq 0)$,

i.e., $|a|<\pi$.
Therefore, the maximum value of the integer $a$ is 3.