## Calculus 8th Edition

Let $y=h(x)=1+\cos x$. $y=1$ for every $x=(2k-1)\frac{\pi}{2}$, where $k$ is an integer, because $\cos (2k-1)\frac{\pi}{2}=0$ (cosine of the odd multiple of the right angle is zero). So to multiple values of the argument $x$ corresponds the same value $y=0$ i.e. the function is not one-to-one.