Question 56, Find the value of $a$ that makes $\sin 4 x \cdot \sin 2 x-\sin x \cdot \sin 3 x=a$ have a unique solution in $[0, \pi)$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Question 56, Solution: Organizing the knowledge:
$$
\begin{array}{l}
\sin 4 x \cdot \sin 2 x - \sin x \cdot \sin 3 x \\
= -\frac{1}{2}(\cos 6 x - \cos 2 x) - \left[-\frac{1}{2}(\cos 4 x - \cos 2 x)\right] \\
= \frac{1}{2}(\cos 4 x - \cos 6 x)
\end{array}
$$

Let \( f(x) = \frac{1}{2}(\cos 4 x - \cos 6 x) \), then \( f(\pi - x) = \frac{1}{2}[\cos 4(\pi - x) - \cos 6(\pi - x)] = \frac{1}{2}(\cos 4 x - \cos 6 x) \). Therefore, \( f(x) \) is symmetric about the line \( x = \frac{\pi}{2} \). Hence, if \( \sin 4 x \cdot \sin 2 x - \sin x \cdot \sin 3 x = a \) has a solution in \( \left(0, \frac{\pi}{2}\right) \), it must also have a solution in \( \left(\frac{\pi}{2}, \pi\right) \), and thus it cannot have a unique solution. Therefore, the only possible unique solution is \( x = 0 \) or \( x = \frac{\pi}{2} \).

If \( x = 0 \) is the unique solution, then \( a = 0 \). However, when \( a = 0 \), \( x = \frac{\pi}{5} \) is also a solution to \( \sin 4 x \cdot \sin 2 x - \sin x \cdot \sin 3 x = 0 \), which contradicts the assumption that \( x = 0 \) is the unique solution;

If \( x = \frac{\pi}{2} \) is the unique solution, then \( a = \sin 2 \pi \cdot \sin \pi - \sin \frac{\pi}{2} \cdot \sin \frac{3 \pi}{2} = 1 \). Since \( \sin 4 x \cdot \sin 2 x - \sin x \cdot \sin 3 x = \frac{1}{2}(\cos 4 x - \cos 6 x) \leq 1 \), and in the interval \( [0, \pi) \), only \( x = \frac{\pi}{2} \) can achieve 1, meaning that the only unique solution is \( x = \frac{\pi}{2} \).
In summary, when \( a = 1 \), the equation has a unique solution.
Note: The graph of \( f(x) = \frac{1}{2}(\cos 4 x - \cos 6 x) \) is as follows: