Precalculus (6th Edition) Blitzer

$x=5+sin(\frac{y}{2})$
Step 1. Given $y=2sin^{-1}(x-5)$, we can determine the domain as $-1\leq x-5\leq 1$ or $4\leq x\leq 6$ Step 2. As $sin^{-1}(x-5)\in [-\frac{\pi}{2},\frac{\pi}{2}]$, we have $y\in [-\pi,\pi]$ Step 3. Rewrite the equation to get $sin^{-1}(x-5)=\frac{y}{2}$ and $x-5=sin(\frac{y}{2})$ Step 4. Thus we have $x=5+sin(\frac{y}{2})$ with $-\pi\leq y\leq\pi$