Example 4 Find the number of solutions to $\sin x=\lg x$.
Analysis This is a transcendental equation, which cannot be solved using elementary methods (thus making it impossible to determine the number of solutions directly).
Let $y_{1}=\sin x=f_{1}(x), y_{2}=\lg x=f_{2}(x)$, then the number of solutions to the original equation is converted to the number of intersection points of the graphs of the two functions $f_{1}(x)$ and $f_{2}(x)$. The discussion of the existence and number of intersection points within a certain range relies on the algebraic analysis of the properties of the functions.
Let $y_{1}=f_{1}(x)=\sin x, y_{2}=f_{2}(x)=\lg x$. Draw the graphs of these two functions in the same Cartesian coordinate system (as shown in Figure 4-1-3, considering only the part where $x>0$ due to the domain restriction).

Figure 4-1-3

From the graph, it can be seen that $y_{1}=\sin x$ and $y_{2}=\lg x$ each have one intersection point in the intervals $\left(\frac{\pi}{2}, \pi\right),\left(2 \pi, \frac{5}{2} \pi\right),\left(\frac{5}{2} \pi, 3 \pi\right)$, and do not intersect in other intervals. If Example 4 is a fill-in-the-blank or multiple-choice question, the answer is obtained up to this point. However, for a complete solution to Example 4, we need to rely on a discussion of the properties of the functions. Let $y=F(x)=y_{1}-y_{2}=\sin x-\lg x$. We will discuss the interval $(0, \pi]$ as an example.

When $x \in(0,1)$, $y_{1}=\sin x>0, y_{2}=\lg x<0$, so $y=F(x)>0$, hence the graph of $y=F(x)$ has no intersection points with the x-axis in $(0,1)$;
When $x \in\left[1, \frac{\pi}{2}\right]$,
$$
\begin{array}{l}
y_{1}=\sin x \geqslant \sin 1>\sin \left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}, \\
y_{2}=\lg x \leqslant \lg \frac{\pi}{2}<0, \\
\text { so } y=F(x)>0, \text { hence the graph of } y=F(x) \text { has no intersection points with the x-axis in } \left[1, \frac{\pi}{2}\right];
\end{array}
$$
When $x \in\left(\frac{\pi}{2}, \pi\right]$, $y_{1}=\sin x$ is a monotonically decreasing function, and $y_{2}=\lg x$ is a monotonically increasing function,
so $y=F(x)=\sin x-\lg x$ is a monotonically decreasing function,
$$
\begin{array}{c}
\text { and } F\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}-\lg \frac{\pi}{2}=1-\lg \frac{\pi}{2}>0, \\
F(\pi)=\sin \pi-\lg \pi=-\lg \pi<0,
\end{array}
$$
so the graph of $y=F(x)$ has exactly one intersection point with the x-axis in $\left(\frac{\pi}{2}, \pi\right]$.

When $x>100$, $\lg x>\lg 100=2 \geqslant 2 \sin x$, so for $x>100$, the graph of the function $y=\lg x$ is always above the graph of the function $y=2 \sin x$. When $x \in[1,100]$, the graphs of these two functions always intersect. When $x \in(100,+\infty)$, the graphs of these two functions do not intersect. Therefore, we only need to find the intersection points of the graphs of these two functions in $x \in[1,100]$. According to the periodicity of the sine function, the number of intersection points is 31.

In general, for a given positive integer $k$, the number of roots of the equation
$$
k \sin x=\lg x
$$
is an unsolved problem, left for interested readers to research and solve.