Trigonometric Inequality
By Hal 2001 on Thursday, September 20,
2001 - 09:23 am:
Find the range of q such that:
-1 < 2sin2 q < 1
I have tried the following: replace the 2sin2 q with 1 - cos2q
then solve as normal...
1 - cos2q cos2q < 2 - - - can't solve, right?
1 - cos2q < 1 0 < cos2q arccos(0) = p/2 pn ±p/4 < q
I doubt that I have answered the question fully and sufficiently?
What would the final inequality look like?
By Olof Sisask on Thursday, September 20,
2001 - 01:16 pm:
You've pretty much got it. Note that
'cos2q < 2 - - - can't solve', isn't quite right, since cos2q < 2
for all q. This means that you don't need to worry about this part of
the inequality.
Then you have
0 < cos2q
Think of the cos graph here, and you'll see that cosx > 0 for
-p/2 < x < p/2 (and also adding 2p to x will not change this, as
this is the same as changing the position of the y-axis).
Therefore,
2q = s+2np, where p/2 < s < p/2 and n is any integer
So, q = t+np with p/4 < t < p/4
Regards,
Olof