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. $\textbf{(A)}\ 3\qquad$ $ \textbf{(B)}\ 4\qquad$ $ \textbf{(C)}\ 5\qquad$ $ \textbf{(D)}\ 6\qquad$ $ \textbf{(E)}\ 7$
Solution
Since $\cos^{-1}(\cos x)=x$ for $x\in[0,\pi]$, and $\arcsin(\sin\theta)\in[-\pi/2,\pi/2]$, solutions require $x\in[0,\pi/2]$.
Writing $\theta=6x\in[0,3\pi]$ and using the piecewise-linear form of $\arcsin(\sin\theta)$ on each length-$\pi$ segment gives the equations $6x=x$ (i.e. $x=0$), $\pi-6x=x$ (i.e. $x=\frac\pi7$), $6x-2\pi=x$ (i.e. $x=\frac{2\pi}5$), $3\pi-6x=x$ (i.e. $x=\frac{3\pi}7$).
Each value lies in its own piece ($6x=0$, $\frac{6\pi}7\in[\frac\pi2,\frac{3\pi}2]$, $\frac{12\pi}5\in[\frac{3\pi}2,\frac{5\pi}2]$, $\frac{18\pi}7\in[\frac{5\pi}2,3\pi]$), so all four are genuine solutions in $[0,\pi/2]$, giving (B) $4$ solutions.
