2021 AMC 12A Spring Problema 19

Question

¿Cuántas soluciones tiene la ecuación

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

en el intervalo cerrado

[0, \pi]

?

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

Respuesta: C

Conceptos:identidad trigonométrica trigonometría

Nivel de dificultad: 2300

Pista pequeña:

Reescribe el lado derecho usando

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

.

Rewrite the right side using

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

Pista grande:

Senos iguales dan

\cos x + \sin x = 1

o

\cos x - \sin x = 1;

cada uno tiene rango

[-\sqrt2, \sqrt2]

.

Equal sines give

\cos x + \sin x = 1

or

\cos x - \sin x = 1;

each has range

[-\sqrt2, \sqrt2]

Solución:

Escribe el lado derecho como

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

. Senos iguales requieren o bien

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

o bien

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

La primera se reduce a

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

; como

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

, solo

k = 0

funciona, dando

\cos x + \sin x = 1

, con soluciones

x = 0

y

x = \tfrac{\pi}{2}

en

[0, \pi]

. La segunda se reduce a

\cos x - \sin x = 1

, cuya única solución en

[0, \pi]

es

x = 0

.

Las soluciones distintas son

x = 0

y

x = \tfrac{\pi}{2}

, para un total de

2

.

Por lo tanto, la respuesta correcta es C.

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.