$8 \cdot 43$ If $\sin 2 x \sin 3 x=\cos 2 x \cos 3 x$, then one value of $x$ is
(A) $18^{\circ}$.
(B) $30^{\circ}$.
(C) $36^{\circ}$.
(D) $45^{\circ}$.
(E) $60^{\circ}$.
(35th American High School Mathematics Examination, 1984)
[Solution 1] From $\sin 2 x \sin 3 x=\cos 2 x \cos 3 x$,
we have
that is $\square$
thus
or
$$
\begin{array}{c}
\cos 2 x \cos 3 x-\sin 2 x \sin 3 x=0, \\
\cos (2 x+3 x)=0 . \\
5 x=90^{\circ}+k \cdot 180^{\circ}, \quad k=0, \pm 1, \pm 2, \cdots \\
x=18^{\circ}+k \cdot 36^{\circ}, \quad k=0, \pm 1, \pm 2, \cdots
\end{array}
$$

Therefore, among the five answers, only $18^{\circ}$ is one of the solutions.
So the answer is (A).
[Solution 2] Observing the original equation, we see that we only need to find the solutions to
$$
\sin 2 x=\cos 3 x \text { and } \sin 3 x=\cos 2 x
$$

Then, these solutions will definitely satisfy the original trigonometric equation. Furthermore, when $2 x$ and $3 x$ are complementary, both equations hold.
Thus, $2 x+3 x=90^{\circ}$, which means $x=18^{\circ}$.
So the answer is $(A)$.