Find the solution set for $[\sin^{-1}x]>[\cos^{-1}x]$, where $[.]$ is greatest integer function The values of $[\sin^{-1}x] \in [-2,1]$ and $[\cos^{-1}x] \in [0,3]$
How do i solve it from here? I have no idea on how to go further. I am not able to convert the graphs to GIFs, and I can’t do it algebraically either.
Note that
$$\lfloor \sin^{-1}x \rfloor=\begin{cases}1, &amp; \sin 1\le x\le 1 \\ 0,&amp; 0\le x\lt \sin 1 \\ -1,&amp; -\sin 1\le x\lt 0 \\ -2,&amp; -1\le x\lt -\sin 1 \end{cases} $$ and
$$\lfloor \cos^{-1}x \rfloor=\begin{cases} 0, &amp;\cos1\lt x\le 1 \\ \vdots\end{cases} $$ We don't need to worry about the other values, as they will turn out to be $\ge 1$, but $\sin^{-1} x\le 1$. Your inequality will only be true, when $$\lfloor \sin^{-1} x\rfloor =1  \land \lfloor \cos^{-1} x\rfloor =0$$ That is, we need to take the intersection of the range of values for which the two equalities hold. The answer is hence $$[\sin 1,1] \cap (\cos 1, 1]\\=\color{purple}{[\sin 1, 1]} \\ (\because \sin 1\gt \cos 1) $$
Not a complete solution
For $x\le0,[\sin^{-1}x]\le-1$  and $[\cos^{-1}x]&gt;\left[\dfrac\pi2\right]=1$
Clearly we need $[\sin^{-1}x],[\cos^{-1}x]\in [0,1]$
$1=[\sin^{-1}x+\cos^{-1}x]\ge[\sin^{-1}x]+[\cos^{-1}x]&gt;2[\cos^{-1}x]$
$\implies [\cos^{-1}x]&lt;\dfrac12$
But $0\le[\cos^{-1}x]\le3\implies[\cos^{-1}x]=0\implies0\le \cos^{-1}x&lt;1\iff x&gt;\cos1\ \ \ \  (1)$
Again  as $[\cos^{-1}]x\ge0,[\sin^{-1}x] $ must be $&gt;0\iff[\sin^{-1}x]\ge1\iff x\ge\sin1\ \ \ \  (2)$
Now $\sin1-\cos1=\sqrt2\sin\left(1-\dfrac\pi4\right)&gt;0,$ so by $(1),(2), x\ge\sin1$ which is obviously a necessary condition
I am trying to find the sufficient condition.