8.47 The number of real roots of the equation $x=\sin x+1993$ is .
(A) 0 .
(B) 1 .
(C) 2 .
(D) More than 2 .
(China Jiangsu Province Suzhou City High School Mathematics Competition, 1993)
[ Sol] Since the roots of the equation are the x-coordinates of the intersection points of the line $y=x$ and the curve $y=\sin x+1993$, the equation has at least one real root.
Assume the equation has two real roots $x_{1}, x_{2}$, and $x_{1} \neq x_{2}$, then
$$
\begin{array}{l}
x_{1}=\sin x_{1}+1993, \quad x_{2}=\sin x_{2}+1993 . \\
\therefore \quad x_{1}-x_{2}=\sin x_{1}-\sin x_{2}=2 \sin \frac{x_{1}-x_{2}}{2} \cos \frac{x_{1}+x_{2}}{2} \text {. } \\
\because \quad\left|\sin \frac{x_{1}-x_{2}}{2}\right|<1, \text{ and } \left|\cos \frac{x_{1}+x_{2}}{2}\right| \leqslant 1 \text{, which is a contradiction.}
\end{array}
$$
Therefore, the original equation has only one real root.
Hence, the correct choice is $(B)$.