Find K knowing tg (x/2)

[darkthorn]

Hello

I am desperate with this homework, I figure out half of the solution but can't find the rest:

Find K, knowing tg (x/2) = 2 (k^2) -k and x belongs to the open interval between pi and 2pi

I manage to find k must be <1/2 but the book solutions say it has to be >0 and I can't understand why.

Thank you

 

[Deleted member 4993]

Hello

I am desperate with this homework, I figure out half of the solution but can't find the rest:

Find K, knowing tg (x/2) = 2 (k^2) -k and x belongs to the open interval between pi and 2pi

I manage to find k must be <1/2 but the book solutions say it has to be >0 and I can't understand why.

Thank you

Please post the EXACT and FULL problem as presented to you.

I cannot make head or tail out of your post!!

You have small "k" and capital "K" - what is the difference between those two?

What is "tg"? What is "tg (x/2)"?

 

[srmichael]

Please post the EXACT and FULL problem as presented to you.

I cannot make head or tail out of your post!!

You have small "k" and capital "K" - what is the difference between those two?

What is "tg"? What is "tg (x/2)"?

I think "tg" means tangent, but I too was confused as to the whole problem.

 

[HallsofIvy]

So you have "tan(x/2)= k^2- k"? That has two unknowns, x and k. You cannot solve a single equation in two unknowns for a single value of each.

You wrote "Knowing K" but there was no "K" in your equations. You were asked if you meant "Knowing k" but did not answer that question. If "K" and "k" are the same thing, then calculate \(\displaystyle k^2- k\), take the arctangent of both sides, and divide both sides by 2.

 

[DrPhil]

Hello

I am desperate with this homework, I figure out half of the solution but can't find the rest:

Find K, knowing tg (x/2) = 2 (k^2) -k and x belongs to the open interval between pi and 2pi

I manage to find k must be <1/2 but the book solutions say it has to be >0 and I can't understand why.

Thank you

\(\displaystyle \pi < x < 2\pi\)

\(\displaystyle \pi/2 < x/2 < \pi \)

\(\displaystyle \tan(x/2) = 2k^2 - k = k\ (2k-1) < 0\), because \(\displaystyle x/2\) is in quadrant II.

\(\displaystyle \implies 0 < k < 1/2\) because the quadratic is negative only between its two roots.