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 $[0, \pi]?$

$0$

$1$

$2$

$3$

$4$

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.

2021 AMC 12A Spring Problema 19

Solución:

El Problema 19 en otros años