2021 AMC 12A Spring Problem 19

Question

How many solutions does the equation

\sin\left(\frac{\pi}{2}\cos x\right) = \cos\left(\frac{\pi}{2}\sin x\right)

have in the closed interval

[0, \pi]?

0

1

2

3

4

Rewrite the right side using

\cos\theta = \sin\!\left(\tfrac{\pi}{2} - \theta\right)

Equal sines give

\cos x + \sin x = 1

or

\cos x - \sin x = 1;

each has range

[-\sqrt2, \sqrt2]

Write the right side as

\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right).

Equal sines require either

\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \frac{\pi}{2}(1 - \sin x) + 2\pi k \end{aligned}

or

\begin{aligned} &\frac{\pi}{2}\cos x \\ &= \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k. \end{aligned}

The first reduces to

\cos x + \sin x = 1 + 4k;

since

\cos x + \sin x \in [-\sqrt2, \sqrt2],

only

k = 0

works, giving

\cos x + \sin x = 1,

with solutions

x = 0

and

x = \tfrac{\pi}{2}

in

[0, \pi].

The second reduces to

\cos x - \sin x = 1,

whose only solution in

[0, \pi]

is

x = 0.

The distinct solutions are

x = 0

and

x = \tfrac{\pi}{2},

for a total of

2.

Thus, the correct answer is C.

Problem 19 in other years