# Trigonometric inequality

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 ($+2n\pi$).

-
Your method is sound. To apply the result to $\cos x$, you have to solve $\cos x \leq -1$ and $\cos x \geq 1/2$ (that is, find the values of $x$ such that $u=\cos x$ lies in $[-1,1]$, for acceptable values of $u$ you have given as intervals, so only $-1$ alone and $[1/2,1]$). The former gives you $-\pi+2k\pi$, the latter is not much more difficult. –  jca May 18 '14 at 9:23
Thank you for the answer. –  tim May 18 '14 at 9:40

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