# Thread: trigonometric functions

1. ## trigonometric functions

I've been working on this problem for awhile now but I think it's easier than I'm making it seem. Here's what I have so far:

3sec^2x = 4
sec^2x = 4/3
secx = sqrt 4/3

It seems right to me but of course sqrt 4/3 isn't a proper answer, since it's not within the domain of sec x. Am I doing this right, and is the answer 'undefined,' or is there something wrong? Please help me, thanks.

2. $\sec^{2} x = \frac{4}{3}$
$\cos^{2} x = \frac{3}{4}$ (by taking the reciprical)
$\cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$

I'm not sure what made you lead to the conclusion that your answer leads to sec x being undefined ..

3. Originally Posted by o_O
$\sec^{2} x = \frac{4}{3}$
$\cos^{2} x = \frac{3}{4}$ (by taking the reciprical)
$\cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$

I'm not sure what made you lead to the conclusion that your answer leads to sec x being undefined ..
Ahh, thank you so much! I put it in my calculator and it gave me an error. My teacher's right, then, when she says we shouldn't depend on them.