Example 4 Solve the equation: $8[3 x]-5[2 x]=3$.

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Let $[3 x]=m,[2 x]=n$.
The original equation becomes $8 m-5 n=3$.
Observation shows that a particular solution is $m=1, n=1$;
The general solution is $m=1-5 t, n=1-8 t(t \in \mathbf{Z})$.
By the property of $[x]$ we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
m \leqslant 3 x<m+1, \\
n \leqslant 2 x<n+1
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\frac{m}{3} \leqslant x<\frac{m+1}{3}, \\
\frac{n}{2} \leqslant x<\frac{n+1}{2} .
\end{array}\right.
\end{array}
$$

Substituting $m, n$ we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{1-5 t}{3} \leqslant x<\frac{2-5 t}{3}, \\
\frac{1-8 t}{2} \leqslant x<\frac{2-8 t}{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\frac{1-5 t}{3}<\frac{2-8 t}{2}, \\
\frac{1-8 t}{2}<\frac{2-5 t}{2}
\end{array}\right. \\
\Rightarrow-\frac{1}{14}<t<\frac{2}{7} .
\end{array}
$$

Since $t$ is an integer, we know $t=0$.
Substituting $t=0$ into the equation set (1) we get
$$
\frac{1}{2} \leqslant x<\frac{2}{3} \text {. }
$$