If sin2xsin3x=cos2xcos3x, then one value for x is

1984 AMC 12 Problem 15

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15.

\sin2x\sin3x=\cos2x\cos3x,

then one value for

x

18^\circ

30^\circ

36^\circ

45^\circ

60^\circ

Move all terms to one side and recognize a cosine addition identity

Use

\cos2x\cos3x-\sin2x\sin3x

=\cos5x

The equation is equivalent to

\begin{aligned} &\cos2x\cos3x\\ &\quad-\sin2x\sin3x\\ &=\cos5x=0. \end{aligned}

Therefore

5x=90^\circ+180^\circ k.

Taking

k=0

gives

x=18^\circ,

the listed value.

Therefore, the correct answer is A.