A simple trigonometric inequality How to sole this inequalities:
$$2\cos\left(\frac{x}{2}\right) \leq -\cos(x).$$
I use the equality $$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos x}{2}}$$ 
to find the solution for the equality 
$$2\cos\left(\frac{x}{2}\right) = -\cos(x).$$
I get a second degree equation $t^2-2t-2=0$ where $t=\cos x$. I think that the solution of the equality are: $$x=\arccos(1-\sqrt3)+2k\pi \\ x=2\pi -\arccos(1-\sqrt3)+2k\pi.$$
How can I find the final solution of the inequality?
 A: Let $\cos\frac{x}{2}=t$.
Thus, we need to solve
$$2t\leq-(2t^2-1)$$
A: \begin{align}
2\cos\frac{x}{2}&\le-\cos x\\
\cos x+2\cos\frac{x}{2}&\le0\\
2\cos^2\frac{x}{2}-1+2\cos\frac{x}{2}&\le0\\
\left(2\cos\frac{x}{2}+1\right)^2&\le3\\
\frac{-1-\sqrt{3}}{2}\le\cos\frac{x}{2}&\le\frac{-1+\sqrt{3}}{2}
\end{align}
So, $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right),4(n+1)\pi-2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right)\right]$.

Note 1:
Consider the equation $\displaystyle 2\cos\frac{x}{2}=-\cos x$. If we prefer to write it as an equation in $\cos x$, we have to square both sides.
\begin{align}
4\cos^2\frac{x}{2}&=\cos^2x\\
2\cos x+2&=\cos^2x\\
\cos^2x-2\cos x-2&=0
\end{align}
The roots are $2n\pi\pm\arccos(1-\sqrt{3})$.
But as squaring the equation $\displaystyle 2\cos\frac{x}{2}=\cos x$ will also yield $\displaystyle 4\cos^2\frac{x}{2}=\cos^2x$, The general solution $x=2n\pi\pm\arccos(1-\sqrt{3})$ includes solutions to both the equations $\displaystyle 2\cos\frac{x}{2}=-\cos x$ and $\displaystyle 2\cos\frac{x}{2}=\cos x$.

Note 2:
For the inequality, as $\displaystyle 2\cos\frac{x}{2}+\cos x$ has a period $4\pi$, we can consider those $x\in[0,4\pi]$ only.
(a) When $x\in[0,\frac{\pi}{2}]$, $\cos\frac{x}{2}>0$ and $-\cos x\le0$. The inequality has no solution.
(b) When $x\in[\frac{\pi}{2},\pi]$, $\cos\frac{x}{2}\ge0$ and $-\cos x\ge0$. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
4\cos^2\frac{x}{2}&\le\cos^2x\\
2\cos x+2&\le\cos^2x\\
\cos^2x-2\cos x-2&\ge0\\
\cos x&\le 1-\sqrt{3}\\
\arccos(1-\sqrt{3})&\le x\le \pi
\end{align}
(c) When $x\in[\pi,\frac{3\pi}{2}]$, $\cos\frac{x}{2}\le0$ and $-\cos x\ge0$. The inequality always holds.
(d) When $x\in[\frac{3\pi}{2},\frac{5\pi}{2}]$, $\cos\frac{x}{2}\le0$ and $-\cos x\le0$. The inequality always holds. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
-2\cos\frac{x}{2}&\ge\cos x\\
4\cos^2\frac{x}{2}&\ge\cos^2x\\
2\cos x+2&\ge\cos^2x\\
\cos^2x-2\cos x-2&\le0\\
1-\sqrt{3}\le\cos x&\le 1+\sqrt{3}\\
\end{align}
which is always true if $x\in[\frac{3\pi}{2},2\pi]$.
(f) When $x\in[\frac{5\pi}{2},3\pi]$, $\cos\frac{x}{2}\le0$ and $-\cos x\ge0$. The inequality always holds.
(g) When $x\in[3\pi,\frac{7\pi}{2}]$, $\cos\frac{x}{2}\ge0$ and $-\cos x\ge0$. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
4\cos^2\frac{x}{2}&\le\cos^2x\\
2\cos x+2&\le\cos^2x\\
\cos^2x-2\cos x-2&\ge0\\
\cos x&\le 1-\sqrt{3}\\
3\pi\le x &\le 4\pi-\arccos(1-\sqrt{3})
\end{align}
(h) When $x\in[\frac{7\pi}{2},4\pi]$, $\cos\frac{x}{2}>0$ and $-\cos x\le0$. The inequality has no solution.
Combining the results in all cases, $\displaystyle \arccos(1-\sqrt{3})\le x\le 4\pi-\arccos(1-\sqrt{3})$.
The general solution is $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+\arccos(1-\sqrt{3}),4(n+1)\pi-\arccos(1-\sqrt{3})\right]$.
A: Directly solve the inequalities with what you know on quadratic polynomials: since the roots of $2t^2+t-1$ are $-1$ and $\dfrac12$,
$$2t^2+t-1\le0 \iff -1\le t\le\frac12\iff \cos\pi\le\cos\frac x2\le \cos\frac\pi3.$$
This means $\;\dfrac\pi3\le\dfrac x2\le\pi\mod2\pi$, hence
$$\dfrac{2\pi}3\le x \le 2\pi\mod 4\pi.$$
