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Trigonometric Inequality
By Hal 2001 on Thursday, September 20, 2001 - 09:23 am:

Find the range of q such that:

-1 < 2sin2q < 1

I have tried the following: replace the 2sin2q 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 +(or)- 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