OKay, so how would this equation be solved,
cosx-2sinxcosx=0
so I've tried this way...
cosx-2sinxcosx=0
cosx(1-2sinx)=0
and after that, i'm stuck.
Well this has multiple answers now doesn't it? You've factored correctly. You know that you're equation will be zero when cos(x)=0 and when 1-2sin(x)=0. So solve for both. I'll start you. The first time cos(x)=0 starting from 0 and moving positively is at $\displaystyle \frac{\pi}{2}$ and it will do so at every at every $\displaystyle \pi$ after that. So I would write one zero as $\displaystyle \frac{\pi}{2}+k\pi$ where k is any integer.