## Calculus 8th Edition

Let $y=g(x)=1-\sin x$. This function is not one-to-one because $y=1$ for every $x$ of the form of $x=k\pi$, where $k$ is integer because $\sin k\pi=0$. This means that to the same value of the function correspond multiple values of the argument so the function is not one-to-one.