If $x,y∈(-π,π]$, then find the area of the polygon formed by points $(x,y)$ satisfying the equation $\lfloor|\sin x|\rfloor+\lfloor|\cos y|\rfloor=2$.
My attempts include using a graphing tool and taking $\sin x= \pm1$ and $\cos y=\pm1$ for $x=\pm \displaystyle{\frac{\pi}{2}}$ and $y=0,\pi$. How do you solve this further? $\lfloor\,\cdot\,\rfloor$ represents the greatest integer function / floor function.
You are on the right track. You know that the equation is satisfied for $x \in \{-\frac{\pi}{2},\frac{\pi}{2}\}$ and $y \in \{0, \pi\}$. So if you take the cartesian product of those two sets you have the list of all the points satisfying this equation : $\{(-\frac{\pi}{2}, 0),\; (-\frac{\pi}{2}, \pi),\;(\frac{\pi}{2}, 0),\; (\frac{\pi}{2}, \pi)\}$. You can see by graphing, or simply by looking at the coordinates that this forms a square of side $\pi$. So the area you are looking for is $\pi^2$.