3. $f(x)$ is an odd function, $g(x)$ is an even function, and $f(x)+g(x)+f(x) g(x)=\left(e^{x}+\sin x\right) \cdot \cos x$, then $f(x)$ has $\qquad$ zeros in $[0,4 \pi)$.
Substitute $-x$ into the original equation, and using the odd and even properties, we get
$$
-f(x)+g(x)-f(x) g(x)=\left(e^{-x}-\sin x\right) \cos x
$$

Adding this to the original equation, we obtain $g(x)=\frac{e^{x}+e^{-x}}{2} \cos x$. When $f(x)=0$, it must be that $g(x)=\left(e^{x}+\sin x\right) \cdot \cos x$, at which point
$$
\left(\frac{e^{x}-e^{-x}}{2}+\sin x\right) \cos x=0,
$$

Noting that $\frac{e^{x}-e^{-x}}{2}+\sin x$ is always greater than 0 for $x \in (0, 4\pi)$, and equals 0 at $x=0$. The solutions to $\cos x=0$ are $x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}$. Therefore, $x$ has 5 possible values: $0, \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}$.

Furthermore, at these values, $g(x)+1 \neq 0$, so $f(x)(g(x)+1)=0$ implies that $f(x)$ must be 0. In summary, $f(x)$ has 5 zeros in $[0, 4\pi)$.