Find the solution of $\lfloor \sin^{-1}(x)\rfloor>\lfloor \cos^{-1}(x)\rfloor$ My question: Find the solution set of $$\lfloor \sin^{-1}(x)\rfloor>\lfloor \cos^{-1}(x)\rfloor$$
Can anyone help me to solve this question? Graphically it seems to be more complicated.
 A: Noting that $\lfloor\arcsin x\rfloor=-2,-1,0,1$ and that $\lfloor\arccos x\rfloor=0,1,2,3$, we have
$$\begin{align}\lfloor\arcsin x\rfloor\gt \lfloor\arccos x\rfloor &\iff \lfloor\arcsin x\rfloor=1\quad\text{and}\quad \lfloor \arccos x\rfloor=0\\&\iff 1\le\arcsin x\lt 2\quad\text{and}\quad 0\le \arccos x\lt 1\\&\iff \sin 1\le x\le 1\quad \text{and}\quad \cos 1\lt x\le 1\\&\iff \color{red}{\sin 1\le x\le 1}\end{align}$$
A: Sure, I'm going to assume you have those floors there on purpose. If so, we know that the inverse trigonometric functions are only defined between $ -1 < x < 1 $. We also know that ${-\pi\over 2}<\sin^{-1}x < {\pi\over 2}$ and ${0}<\cos^{-1}x < {\pi}$. 
Solving the Continuous Problem
Now we can break this up into two regions of interest:


*

*$\cos^{-1}x > \sin^{-1}x \qquad -1 \le x<{1\over \sqrt 2}$

*$\cos^{-1}x < \sin^{-1}x \qquad {1\over \sqrt 2} < x < 1$


We determined that $1 \over \sqrt 2$ was the intersection point by considering the fact that sin and cos are the same when $\theta = {\pi \over 4}$, so we know that ${\pi \over 4}$ is our point of intersect.
Including Floors
To make this discrete, we realise that there is a small region where the two graphs will overlap, when $cos^{-1}$ is floored to 0, this should be evident in the graphic below:

So we therefore just need to solve for $\sin^{-1}x > 1 $ to find the region where it is greater than the inverse cos function. This is accomplished easily by taking the sin of both sides, so the solution to your original question is:

$ x > \sin x $

Hope that helps.
