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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!

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  • $\begingroup$ I think you need to use the fact that $|sin(x)| \leq 1$ for each $x \in \mathbb{R}$. $\endgroup$ – richitesenpai Jun 11 '16 at 1:42
  • $\begingroup$ That struck me at first too, but I couldn't understand how to apply it here. $\endgroup$ – user342209 Jun 11 '16 at 1:43
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    $\begingroup$ The right side is always an integer, while the only integer values the left side could take are -1, 0, and 1. Couldn't you just try each of those? $\endgroup$ – f'' Jun 11 '16 at 1:54
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Continue your idea, we get $$\sin x-[\sin x]=2+[-\cos x].$$ Note that $0\leqslant \sin x-[\sin x]<1$. But, $-1\leqslant -\cos x\leqslant 1$, thus $ [-\cos x]\in\{-1,0,1\}\Rightarrow 2+[-\cos x]\in\{1,2,3\}\Rightarrow 2+[-\cos x]\geqslant 1>x-[\sin x].$

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Note that $\sin(x)$ is an integer. But then $\lfloor \sin(x) \rfloor=\sin(x)$, so your equation reduces to $$\lceil \cos(x) \rceil=2$$ which is not possible.

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