Value of $x$ in floor Inequality: $\lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor$ 
Find value of $x$ for which $\displaystyle \lfloor \sin^{-1}(x)\rfloor >\lfloor \cos^{-1}(x)\rfloor,$ is 

What I try $\displaystyle \sin^{-1}(x)\in \bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$ and $\displaystyle \cos^{-1}(x)\in [0,\pi]$
$\displaystyle \lfloor \sin^{-1}(x)\rfloor \in \{-2,-1,0,1,2\}$ and $\lfloor \cos^{-1}(x)\rfloor \in \{0,1,2,3\}$
$\bullet\; $ If $\lfloor \sin^{-1}(x)\rfloor =2\Rightarrow 2\leq \sin^{-1}(x)<3\Rightarrow \sin 2<x<\sin 3$
Then $\lfloor \cos^{-1}(x)\rfloor =0\Rightarrow 0\leq \cos^{-1}(x)<1\Rightarrow \cos 1<x\leq 1$
$\bullet\ \bullet  $ If $\lfloor \sin^{-1}(x)\rfloor =2\Rightarrow 2\leq \sin^{-1}(x)<3\Rightarrow \sin 2<x<\sin 3$
Then $\lfloor \cos^{-1}(x)\rfloor =1\Rightarrow 1\leq \cos^{-1}(x)<2\Rightarrow \cos 2<x\leq \cos 1$
$\bullet\;\bullet\bullet  $ If $\lfloor \sin^{-1}(x)\rfloor =1\Rightarrow 1\leq \sin^{-1}(x)<2\Rightarrow \sin 1<x<\sin 2$
Then $\lfloor \cos^{-1}(x)\rfloor =0\Rightarrow 0\leq \cos^{-1}(x)<1\Rightarrow \cos 1<x\leq 1$
How do I find common solution Help me please 

$$\lfloor\arcsin x\rfloor>\lfloor\arccos x\rfloor\iff\arcsin x\ge\lfloor\arccos x\rfloor+1.$$
As the arc sine cannot exceed $\dfrac\pi2$ and $\arccos x\ge0$, we have the only case
$$\lfloor\arccos x\rfloor=0\to x>\cos1\land \arcsin x\ge 1\implies x\ge\sin 1.$$
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