Hi Juriemagic and Catmg ,
\(sin\theta=\frac{-6k}{k^2+9}\)
0
The denominator will always be positive, the numerator will always be negative so sin(theta) will always be negative.
We are told that sin(theta) is in the 2nd or third quadrant. Since it is negative it must be in the 3rd quadrant.
Cos(theta) is also negative in the 3rd quadrant.
By inspection I am happy that the value of sin(theta) is between 0 and 1. I can expand on this if you want.
So theta can be any value in the 3rd quadrant [180, 270)
\(sin^2\theta=\frac{(-6k)^2}{(k^2+9)^2}=\frac{36k^2}{(k^2+9)^2}\\ cos^2\theta=1-sin^2\theta\\ cos^2\theta=1-\frac{36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^2+9)^2-36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^4+18k^2+81)-36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^4-18k^2+81)}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^2-9)^2}{(k^2+9)^2}\\ cos\theta=\frac{\pm(k^2-9)}{(k^2+9)}\\ \)
Now 0
\(cos\theta=\frac{k^2-9}{k^2+9}\\\)
Catmg got the second one correct too. Great work Catmg.
To be honest I usually do these more visually.
I'll show you the pic I would normally use:
I'd just read all the ratios off my pic.
Edit: some has not displayed properly ... problems with LaTex. Hopefully it won't matter
Ask questions if you want though. 
Coding:
sin^2\theta=\frac{(-6k)^2}{(k^2+9)^2}=\frac{36k^2}{(k^2+9)^2}\\
cos^2\theta=1-sin^2\theta\\
cos^2\theta=1-\frac{36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^2+9)^2-36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^4+18k^2+81)-36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^4-18k^2+81)}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^2-9)^2}{(k^2+9)^2}\\
cos\theta=\frac{\pm(k^2-9)}{(k^2+9)}\\
cos\theta=\frac{k^2-9}{k^2+9}\\
