2009 AMC 12B 第 24 题

On $[0, \pi],\ \cos^{-1}(\cos x) = x,$ and the left side lies in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ so any solution needs $x \in [0, \tfrac{\pi}{2}]$

Since $\sin$ is one-to-one on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}],$ the equation reduces to $\sin 6x = \sin x$ on $[0, \tfrac{\pi}{2}]$

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 $\sin 6x = \sin x.$

The equation $\sin6x=\sin x$ holds exactly when $$\begin{aligned} 6x&=x+2k\pi, \\ \text{or}\qquad 6x&=\pi-x+2k\pi \end{aligned}$$ for some integer $k.$

The first family gives $x=\tfrac{2k\pi}{5},$ contributing $0$ and $\tfrac{2\pi}{5}.$ The second gives $x=\tfrac{(2k+1)\pi}{7},$ contributing $\tfrac\pi7$ and $\tfrac{3\pi}{7}.$ Hence there are $4$ solutions in $[0,\tfrac\pi2].$

Thus, the correct answer is B.