Example 1 The real solutions of the equation $[2 x]+[3 x]=8 x-\frac{1}{2}$ are $\qquad$
(2006, "Xinzhi Cup" Shanghai Junior High School Mathematics Competition)
Solve: From $[x] \leqslant x<[x]+1$, we get
$$
\begin{array}{l}
2 x-1<[2 x] \leqslant 2 x, \\
3 x-1<[3 x] \leqslant 3 x .
\end{array}
$$

Combining with the original equation, we get
$$
\begin{array}{l}
5 x-2<8 x-\frac{7}{2} \leqslant 5 x \\
\Rightarrow \frac{1}{2}<x \leqslant \frac{7}{6} \Rightarrow \frac{1}{2}<8 x-\frac{7}{2} \leqslant \frac{35}{6} . \\
\text { Also, } 8 x-\frac{7}{2}=[2 x]+[3 x] \text { is an integer, so } \\
8 x-\frac{7}{2}=1,2,3,4,5 .
\end{array}
$$

Solving each, we get $x=\frac{9}{16}, \frac{11}{16}, \frac{13}{16}, \frac{15}{16}, \frac{17}{16}$.
Upon verification, only $x=\frac{13}{16}, \frac{17}{16}$ are solutions to the original equation.