4. 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
4.C.

From the given, we have
$$
\begin{array}{l}
f(x)=\sqrt{13} \sin (x+\varphi)+1, \\
f(x-c)=\sqrt{13} \sin (x+\varphi-c)+1,
\end{array}
$$

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
$$
\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b \sin (x+\varphi-c)+a+b=1 \text {, }
$$

which simplifies to
$$
\sqrt{13}(a+b \cos c) \sin (x+\varphi)-\sqrt{13} b \sin c \cdot \cos (x+\varphi)+(a+b-1)=0 \text {. }
$$

Given the conditions, the above equation must hold for any $x \in \mathbf{R}$, so 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 equations (1) and (3) contradict each other, so $b \neq 0$.
Therefore, from equation (2), we know $\sin c=0$.
When $\cos c=1$, equations (1) and (3) contradict each other, so $\cos c=-1$.
From equations (1) and (3), we find $a=b=\frac{1}{2}$, thus $\frac{b \cos c}{a}=-1$.