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Find the domain of $$f(x) =\lfloor\cos x\rfloor\sin\left(\frac{\pi}{\lfloor x-1\rfloor}\right)$$

where $\lfloor\cdot\rfloor$ is greatest integer function.

Working : Since $\lfloor\cos x\rfloor$ is defined for all real. Only part is the denominator in $\sin\left(\frac{\pi}{\lfloor x-1\rfloor}\right)$ which is $\lfloor x-1\rfloor$ so this part should not be equal to $0$ so that the function should be defined. Also the domain of $\sin x$ is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ Therefore: $-\frac{\pi}{2} \leq \frac{\pi}{\lfloor x-1\rfloor} \leq \frac{\pi}{2}$ Please guide further to solve this..

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    $\begingroup$ The domain of $\sin x$ is $\mathbb{R}$. $\endgroup$ – Javier Mar 8 '13 at 14:51
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The domain of sin is $\mathbb{R}$. So,$[x-1]\neq0$ as you yourself noted. To solve this we note that if $[x-1]=0$, $x\in(0,1]$

So, $\mathbb {R}\setminus(0,1]$ is the domain.

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  • $\begingroup$ but this is not the answer.. $\endgroup$ – Sachin Sharmaa Mar 9 '13 at 8:47

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