I was asked what is the fixed point for the following function? $g(x) = 2sin(\pi x) + x = 0$ between $[1,2]$
because $g(x) = 2sin(\pi x) + x = 0$ doesn't map, I rewrote $g(x)$ as $\frac{1}{\pi}sin^{-1}(\frac{-x}{2})$. However,this choice of g(x) doesn't map either unless I rewrite $g(x)$ as $g(x)=\frac{1}{\pi}(sin^{-1}(\frac{-x}{2})$+$2\pi$). Can someone please tell me where the $+2\pi$ step comes from.
The fixed points of $g(x)=x+ 2 \sin(\pi x)$ are the solutions of the equation $g(x)=x \Leftrightarrow \sin(\pi x) = 0$, which are given by $$ \pi x = k \pi, \quad k\in \mathbb{Z}. $$ So, $g$ has two fixed points in the interval $[1,2]$, namely $x=1$ and $x=2$.