Solve trigonometric inequality $2\cos^2x+\cos(2x) \geq 1$ Can I solve it this way?
$2\cos^2x+\cos^2x-\sin^2x \ge \sin^2x+\cos^2x$
$2\cos^2x-2\sin^2x \ge 0$
$2(\cos^2x-\sin^2x) \ge 0$
$2\cos(2x) \ge 0$
$\cos(2x) \ge 0$
 A: You ended up with $\cos 2x\ge 0\tag 1$
This is possible if and only if $2x$ is in either first or the fourth quadrant. In particular,$(1)$ is true if $2x \in  \color{blue}{[0, \frac \pi 2]\cup [\frac {3\pi} 2,2\pi]}$.
To generalize the blue colored domain, just add $2n\pi, n\in \mathbb Z$ to get the following:
$2x\in [2n\pi, 2n\pi+\frac \pi 2] \cup [ 2n\pi+\frac{3\pi}2, 2n\pi +2\pi], n\in \mathbb Z$.
It follows that $x\in [n\pi, n\pi+\frac \pi 4]\cup [n\pi+\frac{3\pi}4, n\pi +\pi], n\in \mathbb Z$.
A: $$2\cos^2 x + \cos 2x \ge 1\implies 2\cos^2x + 2\cos^2x - 1\ge 1 \implies \cos^2x \ge \frac 12$$
$$\frac 1{\sqrt2} \le \cos x \text{ or }\cos x \le -\frac 1{\sqrt 2}$$
Solving both
$-\frac 1{\sqrt2} \ge \cos x$ and $\cos x \ge\frac 1{\sqrt 2}$
$\color{blue}{x \in \left[n\pi-\frac {\pi}{4}, n\pi+\frac {\pi}{4}\right]}$ where $n\in \mathbb Z$


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*@maths0 comments: using $\cos 2x $
$2\cos^2x  + \cos 2x \ge 1 \implies \cos 2x + 1  + \cos 2x \ge 1 \implies \cos 2x \ge 0$
Instead of solving this problem let's try to understand what exactly it means when $\cos 2x \ge 0$
In short $\cos 2x \ge 0 \implies \text{ cos 2x should be positive and $2x$ must lie in 1st and 4rth quadrant}$
$2x  =  \text{even multiply of }\pi \pm \frac {\pi}{2}$
$x = n\pi \pm \frac {\pi}{4}$
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