8.48 The number of real roots of the equation $\sin x = \lg x$ is
(A) 1.
(B) 2.
(C) 3.
(D) Greater than 3.
(China High School Mathematics League, 1984)
[Solution] In the same coordinate system, draw the graphs of the functions $y_{1}=\sin x, y_{2}=\lg x$ (as shown in the figure) to determine the number of intersections between $y_{1}$ and $y_{2}$, which is the number of real roots of the equation $\sin x=\lg x$.
From the graph, there are three intersection points, meaning the number of real roots of the equation is $3$, so (C) is correct.
In fact, $y_{1}=\sin x$ is a periodic function, and $y_{2}=\lg x$ is an increasing function. We can discuss this by intervals as follows:
Since the domain of $y_{2}=\lg x$ is $x>0$, we only consider the case where $x>0$.
When $x \in(0,1)$, $\sin x>0, \lg x<0$, so in this interval, $y_{1}$ and $y_{2}$ do not intersect, and the original equation has no solution.
When $x \in(1, \frac{\pi}{2})$, $\sin x$ is an increasing function, and $\lg x$ is an increasing function, but $\sin 1>0.8, \lg 1=0, \quad \sin \frac{\pi}{2}=1, \quad \lg \frac{\pi}{2}<1$, so in this interval, $y_{1}$ and $y_{2}$ must intersect at one point.
When $x \in\left(\frac{\pi}{2}, \pi\right)$, $\sin x$ is a decreasing function, and $\lg x$ is an increasing function, but $\sin \frac{\pi}{2}=1, \lg \frac{\pi}{2}<1$, and $\sin \pi=0, \lg \pi>0$, so in this interval, $y_{1}$ and $y_{2}$ must intersect at one point.
When $x \in[\pi, 2 \pi)$, $\sin x \leqslant 0, \lg x>0$, so in this interval, $y_{1}$ and $y_{2}$ do not intersect.
When $x \in\left[2 \pi, \frac{5 \pi}{2}\right)$, $\sin x$ is an increasing function, and $\sin 2 \pi=0, \sin \frac{5 \pi}{2}=1$; while $\lg x$ is an increasing function, and $\lg 2 \pi>0, \quad \lg \frac{5 \pi}{2}>0$, so in this interval, $y_{1}$ and $y_{2}$ intersect at one point.
When $x \in[4 \pi,+\infty)$, $|\sin x| \leqslant 1$, and $\lg x>\lg 10=1$, so in this interval, $y_{1}$ and $y_{2}$ will not intersect again.
In summary, $y_{1}$ and $y_{2}$ have a total of 3 intersections, meaning the original equation has three real roots. Therefore, the answer is (C).