Implicit Function in 2020

Let us solve the equation $x \sin y+\cos y=x$ with respect to $y.$ First, we can solve this equation with respect to $x,$ then we obtain
$x=\frac{\cos y}{1-\sin y}. \;\;\; (*)$ We can transform the right side in the following way: $x= \frac{\cos y}{1-\sin y}= \frac{\sin(\frac{\pi}{2}-y)}{1-\cos(\frac{\pi}{2}-y)}=\frac{2 \sin(\frac{\pi}{4}-\frac{y}{2})\cos(\frac{\pi}{4}-\frac{y}{2})}{2 \sin^2(\frac{\pi}{4}-\frac{y}{2})}=\frac{\cos(\frac{\pi}{4}-\frac{y}{2})}{\sin(\frac{\pi}{4}-\frac{y}{2})}=\frac{1}{\tan(\frac{\pi}{4}-\frac{y}{2})}$
Now, we can solve the last equation for $\tan(\frac{\pi}{4}-\frac{y}{2}),$ then we obtain that $\tan(\frac{\pi}{4}-\frac{y}{2})=\frac{1}{x}$ and solving this equation for $y,$ so we obtain that $y= \frac{\pi}{2}- 2 \arctan(\frac{1}{x}) +2 n \pi$ where $n$ is any integer number. You can notice that this function is discontinuous at $x=0$ because the one sided limits from the left and from the right are different. To construct a function $f(x)$ that is a continuous solution of the given implicit equation on $(-\infty, \infty),$ we need to make $f(x)= \frac{\pi}{2}- 2 \arctan(\frac{1}{x}) +2 n \pi$ for $x>0,$ $f(x)= \frac{\pi}{2}- 2 \arctan(\frac{1}{x}) +2 (n-1) \pi$ for $x<0,$ and $f(0)=-\frac{\pi}{2}+2n\pi.$ Using the condition that $f(1)=2020 \pi,$ we obtain that $n=2020$ and therefore $f(-2)= \frac{\pi}{2}- 2 \arctan(\frac{1}{-2}) +2018 \pi=6342.23...$ Then the answer to the question will be $\boxed{6342.}$

(*)Note: When you divide both sides of the original equation by $1-\sin y$ you are losing the solutions that can be obtained from the equation $\sin y=1.$ These are the functions of the form $f(x)=\frac{\pi}{2}+ 2n\pi$ for any number integer $n.$ None of the graphs of these functions passes through $(1, 2020\pi).$ That is why we don't have to worry about them.

If we write $\alpha(x) = \tan^{-1}x$ , then we see that $\cos\big(f(x) - \alpha(x)\big) \; = \; \tfrac{x}{\sqrt{1+x^2}} \;= \; \sin\alpha(x) \; =\; \cos\big(\tfrac12\pi - \alpha(x)\big)$ so that $0 \; = \; \cos\big(f(x) - \alpha(x)\big) - \cos\big(\tfrac12\pi - \alpha(x)\big) \; = \; -2\sin\big(\tfrac12f(x) - \tfrac14\pi\big) \sin\big(\tfrac12f(x) + \tfrac14\pi - \alpha(x)\big)$ and hence either $f(x) \; =\; \tfrac12\pi + 2n\pi \hspace{2cm} n \in \mathbb{Z}$ or else $f(x) \; = \; 2n\pi - \tfrac12\pi + 2\alpha(x) \hspace{2cm} n \in \mathbb{Z}$ Since $f(1) = 2020\pi$ and $f$ is continuous on $\mathbb{R}$ we deduce that $f(x) \; =\; 2020\pi - \tfrac12\pi + 2\alpha(x) \; =\; 2020\pi - \tfrac12\pi + 2\tan^{-1}x \hspace{2cm} x \in \mathbb{R}$ Putting $x = -2$ we deduce that $\lfloor f(-2)\rfloor = \boxed{6432}$ .

Let $f(x) = \theta$ ; then we have:

$\begin{aligned} x & = x \sin \theta + \cos \theta \\ x & = \frac {\cos \theta}{1-\sin \theta} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = \frac {\frac {1-t^2}{1+t^2}}{1-\frac {2t}{1+t^2}} = \frac {1-t^2}{1+t^2-2t} = \frac {1+t}{1-t} \\ & = \frac {1+\tan \frac \theta 2}{1-\tan \frac \theta 2} = \tan \left(\frac {f(x)}2 + \frac \pi 4\right) \\ \implies f(x) & = 2 (2\blue n\pi + \tan^{-1} x) - \frac \pi 2 & \small \blue{\text{where }n \text{ is an integer.}} \\ f(1) & = 4n\pi + 2 \times \frac \pi 4 - \frac \pi 2 = 2020 \pi & \small \blue{\implies n = 505} \\ \implies f(x) & = 2\tan^{-1} x + 2019.5 \pi \\ f(-2) & = 2\tan^{-1}(-2) + 2019.5\pi \approx 6342.232066 \\ \implies \lfloor f(-2)\rfloor & = \boxed{6342} \end{aligned}$

Let us rewrite the above functional equation according to:

$x \sin f(x) + \sqrt{1-\sin^{2}f(x)} = x$ ;

or $1-\sin^{2}f(x) = x^2[1 - 2\sin f(x) + \sin^{2}f(x)]$ ;

or $0 = (x^2 + 1)\sin^{2} f(x) -2x^2 \sin f(x) + (x^2 - 1);$

or $\sin f(x) = \frac{2x^2 \pm \sqrt{4x^4 - 4(x^2+1)(x^2-1)}}{2(x^2+1)} = \frac{x^2 \pm \sqrt{x^4 - x^4 + 1}}{x^2+1} = \frac{x^2 \pm 1}{x^2+1} = 1, \frac{x^2-1}{x^2+1}$ ;

or $f(x) = \arcsin(1), \arcsin(\frac{x^2-1}{x^2+1});$

or $f(x) = \frac{\pi}{2}, \arcsin(\frac{x^2-1}{x^2+1});$

Since $f(1) = 2020\pi$ , the former constant function poses a contradiction. This leaves us with the latter function which is satisfied according to $2020\pi = \arcsin (\frac{1^2-1}{1^2+1}).$ We are now interested in computing $f(-2)$ , and if we consider:

$\arcsin(\frac{x^2-1}{x^2+1}) = \arccos(\frac{2x}{x^2+1}) \Rightarrow f(-2) = \arcsin (\frac{3}{5}) = \arccos(-\frac{4}{5})$

we have an angle in the second quadrant of the $xy-$ plane. Thus, $f(-2) = 2019\pi - \arcsin(\frac{3}{5}) \Rightarrow \lfloor 2019\pi - \arcsin(\frac{3}{5}) \rfloor = \boxed{6,342}.$