areacalculusgraphing-functionsintegrationtrigonometry
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.
I think you are misinterpreting the question (Its wording is a bit weird imo). I would word it like this
When $x$ satisfies $$\lfloor 5\sin x\rfloor+\lfloor \cos x\rfloor+6=0,$$ then what values can the function $$f(x)=\sqrt 3\cos x+\sin x$$ obtain for these $x$.
Here is an image showing the situation for the interval $[0,2\pi]$:
In this picture, we have:
And you can solve it as follows:
Step 1: Determine the possible values of $x$: First we assume that $x\in[0,2\pi]$ because everything is periodic.
We need that $\sin x< -\frac45$ and $\cos x< 0$, because then $\lfloor 5\sin x\rfloor=-5$ and $\lfloor \cos x\rfloor=-1$ so the equation is satisfied. Clearly, $\cos x< 0$ gives $\frac\pi2< x<\frac{3\pi}2$ (The orange region). You can also solve $\sin x<-\frac45$ to obtain $2\pi-\sin^{-1}\frac45< x<\pi+\sin^{-1}\frac45$ (the red region).
So we find that $x$ is in the interval $\left(2\pi-\sin^{-1}\frac45,\frac{3\pi}2\right)$ (The green region).
Step 2: Find the values that $f$ takes in this interval. First we look at the value of $f$ at the boundary points $2\pi-\sin^{-1}\frac45$ and $\frac{3\pi}2$. We have that $f\left(2\pi-\sin^{-1}\frac45\right) = -\sqrt 3\sqrt{1-\left(-\frac45\right)^2}-\frac45=-\frac{3\sqrt 3+4}5$ and $f(\frac{3\pi}2)=-1$, so we at least know that the range contains these values.
Now we have to see if the function $f$ has a minimum or maximum in this interval. It does not (we see this in the image), so we are done.
Best Answer
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 $\pi^2$.