Question

For how many values of $x$ in $[0, \pi]$ is $\sin^{-1}(\sin 6x) = \cos^{-1}(\cos x)?$

Note: The functions $\sin^{-1} = \arcsin$ and $\cos^{-1} = \arccos$ denote inverse trigonometric functions.

$3$

$4$

$5$

$6$

$7$

Solution:

On $[0, \pi],\ \cos^{-1}(\cos x) = x.$ Since $\sin^{-1}$ takes values in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ any solution requires $x \in [0, \tfrac{\pi}{2}],$ where the equation becomes $\sin 6x = \sin x.$

As $x$ goes from $0$ to $\tfrac{\pi}{2},$ $\sin 6x$ runs $0 \to 1 \to -1 \to 1 \to 0$ (peaks at $\tfrac{\pi}{12}, \tfrac{5\pi}{12},$ trough at $\tfrac{\pi}{4}$ ), while $\sin x$ increases from $0$ to $1.$

Besides $x = 0,$ the graphs cross once in each of $[\tfrac{\pi}{12}, \tfrac{\pi}{4}],$ $[\tfrac{\pi}{4}, \tfrac{5\pi}{12}],$ and $[\tfrac{5\pi}{12}, \tfrac{\pi}{2}],$ for $4$ solutions in all.

Thus, the correct answer is B.

2009 AMC 12B Problem 24

Solution:

Problem 24 in Other Years