i keep on getting the wrong answer but cant work out where im going wrong
solve: between 0 and 360
3cos2 = 2cos^2
3(2cos^2 - 1) = 2cos^2
6cos^2 - 3 = 2cos^2
4cos^2 - 3 = 0
cos = sqr 3/4
Where is the unknown? What you've posted is mathemetically meaningless.
I assume you mean $\displaystyle 3 \cos (2{\color{red}x}) = 2 \cos^2 {\color{red}x}$.
In which case your last line should be $\displaystyle \cos {\color{red}x} = {\color{red}\pm} \frac{\sqrt{3}}{2}$, that is, $\displaystyle \cos {\color{red}x} = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos x = {\color{red}-} \frac{\sqrt{3}}{2}$.