Solve the inequality $|\arccos(\cos(x))|<\pi/4$ To solve this exercise,
$$|\arccos(\cos(x))|<\pi/4$$
I have thought to apply this condition,
$$|f(x)|<k, \quad k\in \Bbb R^+, \iff -k<f(x)<k$$
Hence,
$$-\frac \pi4<\arccos(\cos(x))<\frac \pi4$$
Being $\arccos\colon [-1,1]\to [0,\pi]$, I can have
$$\cos\left(-\frac \pi4\right)<\cos(\arccos(\cos(x)))<\cos\left(\frac \pi4\right) \iff \frac{\sqrt2}2<\cos(x)<\frac{\sqrt2}2$$
false for all $x\in \mathbb{R}$. Are they correct my steps?
 A: Note $\arccos:[-1,1] \to [0,\pi]$ is 1-1 and onto, so $|\arccos \cos x|<\tfrac{\pi}{4}$ is equivalent to $\arccos\cos x \in [0,\tfrac{\pi}{4})$.
Since $\arccos$ is strictly decreasing on its domain, this is equivalent to $\cos x \in (\cos\tfrac{\pi}{4},\cos 0] = (\tfrac{\sqrt{2}}{2},1]$.
This is true for $x\in(-\tfrac{\pi}{4} + 2\pi k,\tfrac{\pi}{4} + 2\pi k)$ for integers $k$.
A: One of your steps is incorrect.  Taking the cosine of an inequality - in your case a double inequality - does not necessarily respect the direction of the inequality symbol because the cosine is not always nonincreasing or nondecreasing.  Indeed, between $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ the cosine changes from increasing to decreasing.
If you wanted to salvage that step you could break it up into cases:
Case 1. $0\leq\cos(\arccos(\cos(x)))<\cos\left(\frac \pi4\right)$ where the cosine is nondecreasing and so taking the cosine does not change the direction of the inequalities
and
Case 2. $\cos\left(-\frac \pi4\right)<\cos(\arccos(\cos(x)))\leq0$ where the cosine is nonincreasing and so taking the cosine changes the direction of the inequalities.
