Prove that $$\sin x=[1+\sin x]+[1-\cos x]$$ has no solution for $x\in \Bbb R$ where $[x]=\lfloor x\rfloor$
$$$$I reduced the equation into $$\sin x=2+[\sin x]+[-\cos x]$$
From here, I plotted the graphs of $\sin x$ and $-\cos x$, and then plottted the graphs of $[\sin x]$ and $[-\cos x]$. Finally, I split the number line into intervals and some distinct points in/at which I added the respective values of $[\sin x]$ and $[-\cos x]$ thus showed that within those intervals/at those points, there were no values of $x$ for which the original equation was valid.
$$$$I found this method quite long and messy (I usually hate methods which involve graphs as I'm a tad slow at them) and was wondering if a shorter and nicer method exists. If so, then could someone kindly show me the method? Thanks!