Wonderful Implicit Function in 2019!

Very similar solution to Arturo Presa's, but with less rigour

$x+1=\sin (x+1+f(x))$

$x+1=-\sin (x+1+f(x)+π)$ where $\sin x = - \sin (x+π)$

$x+1=-\sin (x+1+f(x)-2019π)$ $\sin x$ has a period of $2πn$

$f(x) = -(x+1) - \arcsin (x+1) + 2019π$ which is the form we desire since upon substitution of $x=1, f(x)=2019π$

$f(x)$ in this form is defined for the domain $-2<x<0$

Upon substitution of $x= \frac{1}{2}$ , $⌊100f(-0.5)⌋= 634185$ ​

First, we should understand that the function $\arcsin (\sin x)$ is defined for any values of $x$ and it is periodic with period $2 \pi.$ It is easy to verify that $\arcsin (\sin x)=x$ for any $x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2} ]$ and that $\arcsin (\sin x)=\pi-x$ for any $x$ in the interval $[\frac{\pi}{2},\frac{3 \pi}{2} ].$ Therefore, $\arcsin (\sin x)=x-2n\pi$ on the interval $[-\frac{\pi}{2}+2n \pi,\frac{\pi}{2}+2n \pi ],$ or $\arcsin (\sin x)=-x+(2n+1)\pi$ on the interval $[\frac{\pi}{2}+2n \pi,\frac{3 \pi}{2}+2n \pi ],$ where $n$ is any integer number.

Now taking arcsine of both sides of the equation $x+1=\sin (x+y+1),$ and solving for $y$ it follows that $y= -x-1+2 n \pi+\arcsin(x+1),\;\; \text{or}\;\;y= -x-1+(2 n+1) \pi-\arcsin(x+1),$ where $n$ is any integer number. Notice that both formulas define continuous function on the interval $[-2, 0].$

It is easy to check that the first of these two formulas won't never satisfy the condition that $f(-1)= 2019 \pi.$ Then we need to use the second formula and from this condition, we obtain that the $n=1009.$ Therefore, $f(x)=-x-1+(2019) \pi-\arcsin(x+1),$ and then $\lfloor 100f(-0.5)\rfloor= \boxed{634185}.$