2021 AMC 12A Problem 19

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.