Trigonometric inequality

Question

I have this inequality: $$2\cos^2x+\cos x-1\geq0$$ If I replace $cosx$ with $u$, the inequality becomes:$$2u^2+u-1\geq0$$ The solutions of $$2u^2+u-1=0$$ are $-1$ and $\frac{1}{2}$. When I draw the graph of this function, I can see where it's greater than $0$, that's in the intervals $(-\infty,-1] \cup [\frac{1}{2},+\infty)$. I don't know how to use this information and apply it to $\cos x$ because my results are messed up. (Solution: $x\in[-\frac{\pi}{3}+2n\pi, \frac{\pi}{3}+2n\pi] \cup (\pi+2n\pi), n\in \Bbb Z$)

Edit: The solution is just to continue finding $cosx$ in these intervals, and find the solutions on a unit circle, which gives an interval $x\in[-\frac{\pi}{3}+2n\pi, \frac{\pi}{3}+2n\pi], n\in \Bbb Z$ and a point, because cosine is limited to $-1$, therefore we include $\pi$, and it's multiplies ($+2n\pi$).

Answer 1

Hint: another approach is to write the left hand side as $$ 2\cos^2(x)+\cos(x)-1=2(\cos(x)+\tfrac14)^2-\tfrac98 $$

Answer 2

The solution is just to continue finding $cosx$ in these intervals, and find the solutions on a unit circle, which gives an interval $x\in[-\frac{\pi}{3}+2n\pi, \frac{\pi}{3}+2n\pi], n\in \Bbb Z$ and a point, because cosine is limited to $-1$, therefore we include $\pi$, and it's multiplies ($+2n\pi$).