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

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

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.

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..