Example 6 Find the smallest positive period of the function $y=4 \sin \left(3 x+\frac{\pi}{4}\right)+3 \cos \left(3 x+\frac{\pi}{4}\right)$.
Solution: Since the smallest positive period of $4 \sin \left(3 x+\frac{\pi}{4}\right)$ is $T_{1}=\frac{2 \pi}{3}$, and the smallest positive period of $3 \cos \left(3 x+\frac{\pi}{4}\right)$ is $T_{2}=\frac{2 \pi}{3}$,
and the least common multiple of $T_{1}$ and $T_{2}$ is $\frac{2 \pi}{3}$,
the smallest positive period of the function $y=4 \sin \left(3 x+\frac{\pi}{4}\right)+3 \cos \left(3 x+\frac{\pi}{4}\right)$ is $\frac{2 \pi}{3}$.
Explanation (1) If two trigonometric functions are periodic functions, their smallest positive periods are $T_{1}$ and $T_{2}$, respectively, and $\frac{T_{1}}{T_{2}}$ is a rational number, then the algebraic sum of these two trigonometric functions is certainly a periodic function, and the least common multiple $T$ of $T_{1}$ and $T_{2}$ is one of its periods. However, $T$ is not necessarily the smallest positive period.
In general, we also have:
If the smallest positive periods of two continuous functions $f_{1}(x)$ and $f_{2}(x)$ defined on $\mathbf{R}$ are $T_{1}=p \alpha$ and $T_{2}=\varphi \alpha\left(p, q \in \mathbf{N}^{*},(p, q)=1 . \alpha\right.$ is a positive real number, and $\left.p \neq q\right)$, then the smallest positive period of $f(x)=f_{1}(x)+f_{2}(x)$ is $\frac{p q \alpha}{n}\left(n \in \mathbf{N}^{*}\right.$, and $\left.(n, q p)=1\right)$.

Proof: If $T^{*}$ is the smallest positive period of $f(x)$, then any positive period of $f(x)$ must be an integer multiple of $T^{*}$ (otherwise, if $T=n T^{*}+r\left(0<r<T^{*}, n \in \mathbf{N}^{*}\right)$ is a positive period of $f(x)$, then for any $x$ in the domain, we have $f(x)=f(x+T)=f\left(x+n T^{*}+r\right)=f(x+r)$, which means $r$ is a period of $f(x)$, contradicting the fact that $T^{*}$ is the smallest positive period of $f(x)$).

It is easy to see that $p q \alpha$ is a positive period of $f(x)$, so $p q \alpha=m T^{*}\left(m \in \mathbf{N}^{*}\right)$, thus $T^{*}=\frac{k p q \alpha}{n}=\frac{p q \alpha}{m}$, so $k m=n, k \mid n$, and since $(k, n)=1$, we have $k=1$, so $T^{*}=\frac{p q \alpha}{n},(n, p q)=1$.
(2) The smallest positive period of the sum of two periodic functions can be found using the least common multiple method. Besides the case of the sum of odd and even functions, there are many other scenarios. Here are two more:
(1) When $A_{1} A_{2} \neq 0$, and $\frac{\omega_{1}}{\omega_{2}}$ is a rational number but not equal to 1, the smallest positive period of $f(x)=A_{1} \sin \left(\omega_{1} x+\varphi_{1}\right)+A_{2} \sin \left(\omega_{2} x+\varphi_{2}\right)$ is the least common multiple of the smallest positive periods of $A_{1} \sin \left(\omega_{1} x+\varphi_{1}\right)$ and $A_{2} \sin \left(\omega_{2} x+\varphi_{2}\right)$.
Proof: This has been proven in the derivative method, so it is omitted.
(2) Let $D=\left\{x \mid x \in \mathbf{R}\right.$, but $\left.x \neq k T_{2}+\varphi, k \in \mathbf{Z}\right\}, f_{1}(x)$ is a continuous function on $\mathbf{R}$ with the smallest positive period $T_{1}$, and $f_{2}(x)(x \in D)$ is a periodic function with the smallest positive period $T_{2}$. If $\frac{T_{1}}{T_{2}}$ is a rational number, then the smallest positive period of $f(x)=f_{1}(x)+f_{2}(x)(x \in D)$ is the least common multiple of $T_{1}$ and $T_{2}$.
Proof: Let the least common multiple of $T_{1}$ and $T_{2}$ be $T^{*}$, then there exist positive integers $p, q$ such that $T^{*}=p T_{1}=q T_{2}$, thus
$$
\begin{aligned}
f\left(x+T^{*}\right) & =f_{1}\left(x+T^{*}\right)+f_{2}\left(x+T^{*}\right)=f_{1}\left(x+p T_{1}\right)+f_{2}\left(x+q T_{2}\right) \\
& =f_{1}(x)+f_{2}(x)=f(x),
\end{aligned}
$$

so $f(x)$ is a periodic function, and $T^{*}$ is one of its periods.
If $T$ is a period of $f(x)$, then for any $x \in D$, we have $x+T \in D$, and $f(x+T)=f(x)$, i.e.,
$$
f_{1}(x+T)+f_{2}(x+T)=f_{1}(x)+f_{2}(x) .
$$

If $x_{0}=k T_{2}+\varphi, k \in \mathbf{Z}$, then there must be $x_{0}-T=k^{\prime} T_{2}+\varphi, k^{\prime} \in \mathbf{Z}$ (otherwise, if $x_{0}-T \in$ $D$, then $\left(x_{0}-T\right)+T=x_{0} \in D$, which contradicts $x_{0}=k T_{2}+\varphi, k \in \mathbf{Z}$, i.e., $x_{0} \in I$).
Thus, $\quad T=x_{0}-\left(x_{0}-T\right)=\left(k-k^{\prime}\right) T_{2}=n T_{2}, n=k-k^{\prime} \in \mathbf{Z}$.
Also, $f_{2}(x+T)=f_{2}\left(x+n T_{2}\right)=f_{2}(x)$. Combining (1), we get that for $x \in D$,
$$
f_{1}(x+T)=f_{1}(x) \text {. }
$$

Next, we prove that for any $x_{0} \in \mathbf{R}-D$, (2) also holds.
Take $x_{n} \in D, n=1,2,3, \cdots$, such that $\lim _{n} x_{n}=x_{n}$, by (2) and the fact that $f_{1}(x)$ is a continuous function on $\mathbf{R}$, we get $f_{1}\left(x_{n}+T\right)=f_{1}\left(x_{n}\right)$, taking the limit on both sides $\lim _{n \rightarrow 1} f_{1}\left(x_{n}+T\right)=\lim _{n \rightarrow 1} f_{1}\left(x_{n}\right)$, i.e., $f_{1}\left(x_{n}+T\right)=f_{1}\left(x_{n}\right)$. By the arbitrariness of $x_{0}$, we know that (2) holds for all $x \in \mathbf{R}-D$. Thus, (2) holds for all $x \in \mathbf{R}$, so $T$ is a period of $f_{1}(x)(x \in \mathbf{R})$, hence $T=m T_{1}, m \in \mathbf{Z}$ but $m \neq 0$.

In summary, $T=m T_{1}=n T_{2}$, so $T=k T^{*}, k \in \mathbf{Z}$ but $k \neq 0$. Therefore, the smallest positive period of $f(x)(x \in D)$ is the least common multiple of $T_{1}$ and $T_{2}$, which is $T^{*}$.