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
Using $\cos\theta=\sin(\pi/2-\theta)$: the equation becomes $\sin\left(\frac\pi2\cos x\right)=\sin\left(\frac\pi2-\frac\pi2\sin x\right)$.
Case A: $\frac\pi2\cos x=\frac\pi2-\frac\pi2\sin x+2k\pi$, i.e. $\cos x+\sin x=1+4k$; since $\cos x+\sin x\in[-\sqrt2,\sqrt2]$, only $k=0$ works, giving $\cos x+\sin x=1$, i.e. $x=0$ or $x=\pi/2$.
Case B: $\sin u=\sin v$ also holds when $u=\pi-v+2k\pi$; this gives $\frac\pi2\cos x=\frac\pi2+\frac\pi2\sin x+2k\pi$, i.e. $\cos x-\sin x=1+4k$; only $k=0$ works, giving $\cos x-\sin x=1$, i.e. $x=0$ (for $x\in[0,\pi]$).
Altogether the solutions are $x=0$ and $x=\pi/2$ (check: $x=\pi/4$ does not work, since $\sin(\frac\pi2\cdot\frac{\sqrt2}2)\ne\cos(\frac\pi2\cdot\frac{\sqrt2}2)$). So there are $2$ solutions, answer (C).
