2021 AMC 12A Problems/Problem 19

Problem

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 $[0,\pi]$ ?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

Solution 1 (Inverse Trigonometric Functions)

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

The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right]$ , which is included in the range of $\arcsin$ , so we can use it with no issues.

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

$\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x$

$\cos x = 1 - \sin x$

$\cos x + \sin x = 1$

This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi]$ , because one of $\sin$ and $\cos$ must be $1$ and the other $0$ . Therefore, the answer is $\boxed{C) 2}$

~Tucker

Solution 2 (Analysis)

Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)

https://youtu.be/wJxN1YPuyCg

~ pi_is_3.14

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2021 AMC 12A Problems/Problem 19

Contents

Problem

Solution 1 (Inverse Trigonometric Functions)

Solution 2 (Analysis)

Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)

See also