Wow, I had no idea about that rule, looks like I shouldn't have passed trig. :/ Haha, thanks everyone who helped me figure this out, very much appreciated. Going to post my final solution because I hate when I find a topic and there's no final sum up of everything to help it all make sense.
pi = the mathematical constant pi in the following equations.
So we are looking for the solution to: 2*sin(pi * x) + x = 0 on the interval [1,2] using Banach's fixed point theorem.
Solving for x, and concluding x = -2*sin(pi * x) will not wield a solution using the algorithm because the derivative is too large near the fixed point ( >= 1.0).
So we need to find another way to solve for x. Dick and haruspex pointed out that we can't use arcsin on the interval we are working with currently, because it will only work on the interval from [-pi/2, pi/2].
So we shift the argument of pi first:
2*sin(pi*x) + x = 2 * -sin(pi*x - pi) + x //uses trigonometric identities
NOW we solve for x:
2*-sin(pi*x - pi) + x = 0 //Initial equation
2*-sin(pi*x - pi) = -x //Bring the x to the right
-sin(pi*x - pi) = -x/2 //Bring the 2 to the right
sin(pi*x - pi) = x/2 //Drop the negative signs
pi*x - pi = arcsin(x/2) //Get rid of sin on the left using arcsin
pi(x - 1) = arcsin(x/2) //Factor out pi from the left
x - 1 = (1/pi) * arcsin(x/2) //Bring the pi to the right
x = (1/pi) * arcsin(x/2) + 1 //Bring the one to the right
so g(x) = (1/pi) * arcsin(x/2) + 1
g'(x) = 1/(pi * (sqrt(4-x^2))) //Used my calculator to solve this
1.) Prove the function g(x) maps from [1, 2] -> [1, 2]
We only need to solve g(1) and g(2) because these will be the min and max values, since there is no solution to the equation g'(x) = 0.
g(1) = 1.1666
g(2) = 1.5
OK, so that's proof.
2.) Prove g(x) is a contraction on [1, 2]
We have to show that maximum value from [1,2] of g'(x) is less than 1.
Oh, what the helllllll? No, I don't care I'm rewriting the range, the book says on that interval but its undefined at 2 because of the divide by zero. The range is now [1, 1.5] because sometimes you have to take math into your own hands.
We have to show that maximum value from [1,1.5] of g'(x) is less than 1.
Well, the MAXIMUM value would have to be at 1.5, because it gives the smalls possible number at the denominator, and that solution is g'(1.5) = .2406, so yes, it works on this range. Above, if you change the range it works too. As practice, try doing it yourself!
OK, so this will work with Banach's Fixed Point algorithm, and the solution is about 1.206035, so haruspex was correct. Woo math, thank you one and all who helped, you're truly gods among men.