Fixed point iteration doubt

Hi! I was asked to find an approximation for a root of on the interval [1, 2] using the fixed point iteration method, with p0=1.

So, I got

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 and got

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

Re: Fixed point iteration doubt

It's because the inverse sine is multivalued. When you attempt to evaluate your calculator will return a negative value whereas what you want is the value that is further on (if that makes sense).

The iteration that you need is

from which

Re: Fixed point iteration doubt