I have a problem to solve this inequality $\sin(2x)-2\cos x\geq 0$

Question

I have a problem to solve this inequality: $$\sin(2x)-2\cos x\geq 0$$

I have tried at similar mode:

$$2\sin x\cos x-2\cos x\geq 0 \Longleftrightarrow 2\cos x(\sin x-1)\geq 0$$

My result is

$$-\frac \pi2+2k\pi\leq x \leq \frac \pi2+2k\pi, \quad k\in\mathbb{Z}$$ because $\cos x\geq 0,-\frac \pi2+2k\pi\leq x \leq \frac \pi2+2k\pi, \quad \text{and} \quad \sin x-1\geq 0\,\,$ if $\,\,\,\,\,x=\pi/2+2k\pi$.

The solution of my book is:

$$\frac \pi2+2k\pi\leq x \leq \frac 32\pi+2k\pi, \quad k\in\mathbb{Z}$$ Why?

Answer 1

You showed that each term is only non-negative if $x=\pi/2+2k\pi$. So this is a solution to the inequality. But you missed one thing - what about if both terms are negative? Then their product is still positive. Therefore this also solves your equation, so if you want a complete solution, you must also solve $$\cos x\le 0,\,\,\,\sin x-1\le0$$

The second of these is true for all $x\in\Bbb R$, the first condition is what will give you the full solution.

Answer 2

Hint: $\sin{x}-1$ is always less or equal to zero isn't it?

What does your $\cos{x}$ need to be then?

Answer 3

As $\sin x-1\le0$, we need $\cos x\le 0$ instead of $\cos x\ge 0$.

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Note that

$2\cos x(\sin x-1)\ge0 \implies \begin{cases} \cos x>0 \\ \sin x-1>0 \end{cases} \textrm{ or }\begin{cases} \cos x<0 \\ \sin x-1<0 \end{cases} \textrm{ or }\cos x=0 \textrm{ or } \sin x-1=0$

As $\sin x -1$ cannot be $<0$, one of the cases can be rejected.