Find the domain of a function of the form: ln(sin(1/x))

[sergey_le]

TL;DR Summary

I know we need to find out when sin(pi/x)>0.
But I can 't do it

the solution

 

[fresh_42]

If you know the sine function, why can't you say when $y=\sin x$ is positive?

 

[sergey_le]

If you know the sine function, why can't you say when $y=\sin x$ is positive?

Because it's not just a sinx.
It's sin(1/x)

 

[fresh_42]

Because it's not just a sinx.
It's sin(1/x)

Proceed step by step. When is $y=\sin x > 0$? Which are the intervals of $x$ for which this is true? Let's change the variable names. When is $Y=\sin X > 0$? If we know that, we get e.g. $X\in (0,\pi)$.

With that we have now $X=\dfrac{\pi}{x}$. That is $0<X=\dfrac{\pi}{x} < \pi$. Can you restructure that in a way, that $a<x<b$ is the result?

 

[sergey_le]

Proceed step by step. When is $y=\sin x > 0$? Which are the intervals of $x$ for which this is true? Let's change the variable names. When is $Y=\sin X > 0$? If we know that, we get e.g. $X\in (0,\pi)$.

With that we have now $X=\dfrac{\pi}{x}$. That is $0<X=\dfrac{\pi}{x} < \pi$. Can you restructure that in a way, that $a<x<b$ is the result?

I know that sinx> 0 when 0+2πk<x<π+2πk.

I do not understand why it is okay to say that 0+2πk<π/x<π+2πk.

 

[fresh_42]

I know that sinx> 0 when 0+2πk<x<π+2πk.

I do not understand why it is okay to say that 0+2πk<π/x<π+2πk.

Let us write the sine function as $\varphi = \sin \alpha$. Then you just said that $\varphi >0$ when $2k\pi < \alpha < (2k+1)\pi$ for $k\in \mathbb{Z}$. Now how can we compare $\sin \alpha$ and $\sin (\pi/x)$? We can do this by setting $\alpha:=\pi/x.$ Thus we get from the above inequality that
$$
\varphi =\sin \alpha = \sin \left(\dfrac{\pi}{x}\right) > 0 \text{ when } 2k\pi < \alpha = \dfrac{\pi}{x} < (2k+1)\pi \quad (k\in \mathbb{Z})
$$
Next question is: Can you transform $2k\pi < \dfrac{\pi}{x} < (2k+1)\pi$ into something like $A < x < B$?

 

[sergey_le]

Let us write the sine function as $\varphi = \sin \alpha$. Then you just said that $\varphi >0$ when $2k\pi < \alpha < (2k+1)\pi$ for $k\in \mathbb{Z}$. Now how can we compare $\sin \alpha$ and $\sin (\pi/x)$? We can do this by setting $\alpha:=\pi/x.$ Thus we get from the above inequality that
$$
\varphi =\sin \alpha = \sin \left(\dfrac{\pi}{x}\right) > 0 \text{ when } 2k\pi < \alpha = \dfrac{\pi}{x} < (2k+1)\pi \quad (k\in \mathbb{Z})
$$
Next question is: Can you transform $2k\pi < \dfrac{\pi}{x} < (2k+1)\pi$ into something like $A < x < B$?

Thank you very much, I finally understood