# Thread: find all solutions for sec^2 (theta) - 4 = 0

1. ## find all solutions for sec^2 (theta) - 4 = 0

okay so I get that I factor it out with difference of two squares

(sec theta - 2)(sec theta +2)=0
so
sec theta = 2
sec theta= -2

therefore
cos theta = 1/2 = pi/3 which gives me the solutions: (pi/3 + 2pi N) and (2pi/3 + 2pi N)

what I don't understand is this...since cos theta also = -1/2 shouldn't the solutions also include

(4pi/3+2pi N) and (5pi/3 + 2pi N)?

2. ## Re: find all solutions for sec^2 (theta) - 4 = 0

$\cos{t} = \frac{1}{2}$ at $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}$

$\cos{t} = -\frac{1}{2}$ at $t = \frac{2\pi}{3}$ and $t = \frac{4\pi}{3}$