trig

**pnfuller** · October 17th 2012, 03:05 PM

when does sec²x equal 4?

**MarkFL** · October 17th 2012, 03:17 PM

Take the square root of both sides, then write in terms of cosine if that helps. What do you find?

**pnfuller** · October 17th 2012, 03:21 PM

pi/3... 2pi/3... 4pi/3... 5pi/3

**MarkFL** · October 17th 2012, 03:24 PM

Yes, assuming we are given $0\le x<2\pi$ .

**pnfuller** · October 17th 2012, 03:26 PM

or plus 2npi for each if not given that interval?

Originally Posted by MarkFL2

Yes, assuming we are given $0\le x<2\pi$ .

**MarkFL** · October 17th 2012, 06:14 PM

We could shorten it since the 1st quadrant solution and the 3rd quadrant solutions differ by $\pi$ , likewise for the 2nd and 4th quadrant roots, and so we could state:

$x=\frac{\pi}{3}(3k+1),\,\frac{\pi}{3}(3k+2)$ where $k\in\mathbb{Z}$ .

**Kyood** · October 27th 2012, 02:07 AM

Originally Posted by MarkFL2

We could shorten it since the 1st quadrant solution and the 3rd quadrant solutions differ by $\pi$ , likewise for the 2nd and 4th quadrant roots, and so we could state:

$x=\frac{\pi}{3}(3k+1),\,\frac{\pi}{3}(3k+2)$ where $k\in\mathbb{Z}$ .

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