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Psychiccleo21 · Dec 8, 2007, 04:38 PM

How do I solve this?

Question: Use inverse functions where needed to find all solutions of the equation in the interval [0, 2 pi)

3tan^2x - 11sec x + 13 = 0

s_cianci · Dec 8, 2007, 04:44 PM

Start by changing everything to sin and cos. Then combine everything into 1 single rational expression. It falls into place from there.

galactus · Dec 8, 2007, 05:29 PM

Another way is to remember that $tan^{2}(x)=sec^{2}(x)-1$

Then we have:

$3(sec^{2}(x)-1)-11sec(x)+13=0$

$3sec^{2}(x)-11sec(x)+10=0$

Now, let $u=sec(x)$

We get: $3u^{2}-11u+10=0$

Now, solve the quadratic.

Psychiccleo21 · Dec 8, 2007, 07:37 PM

Originally Posted by Psychiccleo21

How do i solve this?

Question: Use inverse functions where needed to find all solutions of the equation in the interval [0, 2 pi)

3tan^2x - 11sec x + 13 = 0

I had it figured up to the point where I had cos x = 3/5 and cos x = 1/2, but after that, I had no idea what to do. I know I can find the points where cos = 1/2 by looking on a unit circle, but 3/5..