2. 1. $7^{\star \star}$ Let the function $f(x)=3 \sin x+2 \cos x+1$. If real numbers $a, b, c$ make $a f(x)+$ $b f(x-c)=1$ hold for any real number $x$, then the value of $\frac{b \cos c}{a}$ is ( ).
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) -1
(D) 1
It is known that for any $x \in \mathbf{R}$, we have $f(x) + f(x - \pi) = 2$. Therefore, taking $a = b = \frac{1}{2}$, $c = \pi$, we get that for any $x \in \mathbf{R}$, $a f(x) + b f(x - c) = 1$, from which we derive $\frac{b \cos c}{a} = -1$.
More generally, from the given conditions, we have
$$
f(x) = \sqrt{13} \sin (x + \varphi) + 1, \quad f(x - c) = \sqrt{13} \sin (x + \varphi - c) + 1,
$$

where $0 < \varphi < \frac{\pi}{2}$ and $\tan \varphi = \frac{2}{3}$. Thus, $a f(x) + b f(x - c) = 1$ can be rewritten as
$$
\sqrt{13} a \sin (x + \varphi) + \sqrt{13} b \sin (x + \varphi - c) + a + b = 1,
$$

which simplifies to $\sqrt{13} a \sin (x + \varphi) + \sqrt{13} b \sin (x + \varphi) \cos c - \sqrt{13} b \cos (x + \varphi) \sin c + (a + b - 1) = 0$, so $\sqrt{13}(a + b \cos c) \sin (x + \varphi) - \sqrt{13} b \sin c \cos (x + \varphi) + (a + b - 1) = 0$.
By the given conditions, the above equation holds for any $x \in \mathbf{R}$, hence it must be that $\left\{\begin{array}{l}a + b \cos c = 0, \\ b \sin c = 0, \\ a + b - 1 = 0 .\end{array}\right.$
If $b = 0$, then from (1) we have $a = 0$, which clearly does not satisfy (3). Therefore, $b \neq 0$. From (2), we know $\sin c = 0$, so $c = 2k\pi + \pi$ or $c = 2k\pi (k \in \mathbf{Z})$. When $c = 2k\pi$, $\cos c = 1$, then (1) and (3) are contradictory. Hence, $c = 2k\pi + \pi (k \in \mathbf{Z}) \cos c = -1$. From (1) and (3), we know $a = b = \frac{1}{2}$, so $\frac{b \cos c}{a} = -1$, thus the answer is C.