View Full Version : Inverse functions/ unit circle
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
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..