A simple trigonometric inequality

Question

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?

Answer 1

Let $\cos\frac{x}{2}=t$.

Thus, we need to solve $$2t\leq-(2t^2-1)$$

Answer 2

\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, $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right),4(n+1)\pi-2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right)\right]$.

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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 $\displaystyle 2\cos\frac{x}{2}=\cos x$.

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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 $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+\arccos(1-\sqrt{3}),4(n+1)\pi-\arccos(1-\sqrt{3})\right]$.

Answer 3

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 $$\dfrac{2\pi}3\le x \le 2\pi\mod 4\pi.$$