Example 7 Find the smallest positive period of the function $f(x)=\sin 2 x+2 \cos 2 x+3 \tan 4 x$.
Solve: The domain $D=\left\{x \mid x \in \mathbf{R}\right.$, but $\left.x \neq \frac{1}{4} k \pi+\frac{\pi}{8}, k \in \mathbf{Z}\right\}$,
the function $f_{1}(x)=\sin 2 x+2 \cos 2 x=\sqrt{5} \sin (2 x+\varphi)$, the smallest positive period is $T_{1}=\frac{2 \pi}{2}=\pi, f_{2}(x)$ $=3 \tan 4 x$ the smallest positive period $T_{2}=\frac{\pi}{4}$, hence the smallest positive period of $f(x)$ is the least common multiple of $\pi$ and $\frac{\pi}{4}$, which is $\pi$.

Explanation: Generally, for $f(x)=a \sin ^{n}\left(\omega_{1} x+\varphi_{1}\right)+b \tan ^{m}\left(\omega_{2} x+\varphi_{2}\right)\left(n, m \in \mathbf{N}^{*}, a b \neq 0, \omega_{1} \omega_{2} \neq 0\right)$, the smallest positive period is the least common multiple of the smallest positive periods of the functions $g(x)=\sin ^{n}\left(\omega_{1} x+\varphi_{1}\right)$ and $h(x)=\tan ^{m}\left(\omega_{2} x+\varphi_{2}\right)$, $T_{1}$ (when $n$ is odd, it equals $\frac{2 \pi}{\left|\omega_{1}\right|}$; when $n$ is even, it equals $\frac{\pi}{\left|\omega_{1}\right|}$), and $T_{2}\left(=\frac{\pi}{\left|\omega_{2}\right|}\right)$.