Hi! I was asked to find an approximation for a root of $\displaystyle f(x)=2sin(x\pi)+x=0$ on the interval [1, 2] using the fixed point iteration method, with p0=1.

So, I got $\displaystyle g(x)=x=(\arcsin(-x/2))/\pi$

However, that way I just get numbers very close to zero (which is a solution for the equation, but it's not on the interval I want).

Anyway, I checked the answer on the back of the book and they used $\displaystyle g(x)=x=(\arcsin(-x/2))/\pi+2$ and got $\displaystyle x \simeq 1.68$

I just don't know from where does that +2 come? Could anyone explain me, please?