Inequality

[Loki123]

Is my thought process correct here?

 

[pka]

To Loki, you must tell us the exact wording of the question.
The is no point in our trying to guess from your post.

 

[Loki123]

To Loki, you must tell us the exact wording of the question.
The is no point in our trying to guess from your post.

there is no question. I just wanted to check whether I undestand inequality with trigonometry. As you can see cosx is equal or above 0, bellow is how i would define x then. Same goes for the other. Is how I defined x okay?

 

[Dr.Peterson]

there is no question. I just wanted to check whether I undestand inequality with trigonometry. As you can see cosx is equal or above 0, bellow is how i would define x then. Same goes for the other. Is how I defined x okay?

I think your solutions to the inequalities are all correct, as I read them, except for the last. Some words would have helped.

Look carefully at that last one. What does it say if k=0, for example?

And technically, what you mean in each case is the union of those intervals over all integers k.

 

[Loki123]

I think your solutions to the inequalities are all correct, as I read them, except for the last. Some words would have helped.

Look carefully at that last one. What does it say if k=0, for example?

And technically, what you mean in each case is the union of those intervals over all integers k.

(pi, 0) that still makes sense to me though. What am I missing.

 

[Dr.Peterson]

(pi, 0) that still makes sense to me though. What am I missing.

In interval notation, the second number is always greater than the first. And if that weren't required, (pi, 0) would presumably mean the same as (0, pi), which is not what you mean, is it? You mean (pi, 2 pi), where the sine is negative!

 

[Loki123]

In interval notation, the second number is always greater than the first. And if that weren't required, (pi, 0) would presumably mean the same as (0, pi), which is not what you mean, is it? You mean (pi, 2 pi), where the sine is negative!

So maybe (Pi+2kPi, 2Pi + 2kPi)