Given a map $T(x) = 2 \sin(x)$, find a fixed point (It is on $(\frac{\pi}{2}, \pi)$).
Note that $T(x)$ is a contraction on $(\frac{\pi}{3}, 2\frac{\pi}{3})$. By Banach contraction principle, $T^n(x_0)$ converges to $x^*$ .
How can we approximate $x^*$ up to $4$ decimal places?
I set $x_0 = \frac{\pi}{3}$ and repeat the process initiated by $T(x_0)=2\sin(\frac{\pi}{3})$, and continue this way.
But it won't give $x^*$.
Starting with $\pi/3$ will converge to the fixed point at $x = 0$. Draw the picture for the iterates to makes sure that you understand what's going on here. It's relevant that the function is increasing on $[0, \pi/2]$, and lies strictly above the line $y = x$ here.
Starting with $\pi/2$ will converge to a fixed point near $1.8955$ (accurate to $4$ places) after about $20$ iterations.