Domain of a trig. function!

**javax** · March 23rd 2008, 07:25 AM

Hi guys.
I need help on finding the domain of this function:

$y = \sqrt{sin\frac{1}{x}}$

**Soroban** · March 23rd 2008, 09:37 AM

Hello, javax!

A tricky one . . .

I need help on finding the domain of this function:

. . $y \:= \:\sqrt{\sin\frac{1}{x}}$

We see that: . $\sin\frac{1}{x} \,\geq\,0$

This happens when $\frac{1}{x}$ is on the intervals:

. . $\hdots\;(-4\pi,-3\pi),\;(-2\pi,-\pi),\;(0,\pi),\;(2\pi,3\pi),\;(4\pi,5\pi),\;\hdot s$

That is: . $2n\pi \:\leq\:\frac{1}{x} \:\leq\<img src=$ 2n+1)\pi\;\text{ for any integer }n " alt="2n\pi \:\leq\:\frac{1}{x} \:\leq\

2n+1)\pi\;\text{ for any integer }n " />

Therefore: . $\frac{1}{(2n+1)\pi} \;\leq \;x \;\leq \; \frac{1}{2n\pi}$

(I'll let you explain what happens when $n = 0$ .)

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