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
$$
\begin{array}{l}
\text { According to the problem, } 3 a \sin x+2 a \cos x+a+3 b \sin (x-c)+2 b \cos (x-c)+b \\
=\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b \sin (x-c+\varphi)+a+b \\
=\sqrt{13} a \sin (x+\varphi)+\sqrt{13} b[\sin (x+\varphi) \cos c-\cos (x+\varphi) \sin c]+a+b \\
=\sqrt{13}[(a+b \cos c) \sin (x+\varphi)-b \sin c \cos (x+\varphi)]+a+b \\
=\sqrt{13} \cdot \sqrt{(a+b \cos c)^{2}+(b \sin c)^{2}} \sin (x+\varphi-\theta)+a+b \equiv 1 \\
\Rightarrow\left\{\begin{array} { l } 
{ a + b \operatorname { c o s } c = 0 , } \\
{ b \operatorname { s i n } c = 0 , } \\
{ a + b = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=b=\frac{1}{2}, \\
c=2 k \pi+\pi(k \in \mathbf{Z})
\end{array} \Rightarrow \frac{b \cos c}{a}=-1 . \text { Therefore, the answer is } C .\right.\right.
\end{array}
$$
$$