Solving $\left\lfloor \sin \left( x\right) \right\rfloor -\left\lfloor \sin \left( \left\lfloor x\right\rfloor \right) \right\rfloor =1$

I've created this equation $\left\lfloor \sin \left( x\right) \right\rfloor -\left\lfloor \sin \left( \left\lfloor x\right\rfloor \right) \right\rfloor =1$ for real numbers $x$. I've found a complicated formula for the solutions and I'm not sure about it since I found it hard to simplify it.The solution I've found is :$x\in \left( \underset{k\in\mathbb{Z}}{\bigcup }\left( \left[ 2k\pi ,\left( 1+2k\right) \pi \right] \diagup\left\{ \frac{\pi }{2}+2k\pi \right\} \right) \right) \cap \left( \underset{l\in\mathbb{Z}}{\bigcup }\left[ \left\lfloor \left( -1+2l\right) \pi \right\rfloor+1,\left\lfloor 2l\pi \right\rfloor +1\right[ \right) $

It seems I've missed something, since this I've simplified it to :

$x\in \underset{k\in\mathbb{Z}}{\bigcup }\left[ 2k\pi ,\left\lfloor 2k\pi \right\rfloor +1\right[ $

But this last formula don't show that $\frac{\pi }{2}$ is a solution. So any help please ?