Solving Trigonometric Inequalities: Very Difficult Problems with Solutions

Solution:
Let [tex]s=\sin x[/tex]. Since [tex]\cos 2x=1-2s^2[/tex], the inequality is [tex]7-4s^2\le 3\left|2s-1\right|[/tex].
If [tex]s\ge\frac{1}{2}[/tex]: [tex]7-4s^2\le 6s-3\iff(s-1)(2s+5)\ge 0[/tex], so [tex]s\ge 1[/tex], i.e. [tex]s=1[/tex]. If [tex]s\lt\frac{1}{2}[/tex]: [tex]7-4s^2\le 3-6s\iff(s-2)(2s+1)\ge 0[/tex], so [tex]s\le-\frac{1}{2}[/tex].
Finally, [tex]\sin x=1\iff x=\frac{\pi}{2}+2k\pi[/tex], and [tex]\sin x\le-\frac{1}{2}\iff x\in\left[\frac{7\pi}{6}+2k\pi, \frac{11\pi}{6}+2k\pi\right][/tex].

Solution:
Convert the products to sums:
[tex]\sin x\sin 3x=\frac{1}{2}(\cos 2x-\cos 4x), \quad\sin 5x\sin 7x=\frac{1}{2}(\cos 2x-\cos 12x)[/tex]
The inequality becomes [tex]\cos 12x\gt\cos 4x[/tex], i.e.
[tex]\cos 4x-\cos 12x=2\sin 8x\sin 4x=4\sin ^2 4x\cos 4x\lt 0[/tex]
So [tex]\cos 4x\lt 0[/tex] and [tex]\sin 4x\ne 0[/tex]: [tex]\frac{\pi}{2}+2k\pi\lt 4x\lt\frac{3\pi}{2}+2k\pi[/tex] and [tex]4x\ne\pi+2k\pi[/tex]. Dividing by [tex]4[/tex]: [tex]\frac{\pi}{8}+\frac{k\pi}{2}\lt x\lt\frac{3\pi}{8}+\frac{k\pi}{2}[/tex] and [tex]x\ne\frac{\pi}{4}+\frac{k\pi}{2}[/tex].

Solution:
All sides must be positive. [tex]\sin x\gt 0[/tex] and [tex]\sin 2x\gt 0[/tex] give [tex]x\in\left(2k\pi, \frac{\pi}{2}+2k\pi\right)[/tex]; then [tex]\sin 3x\gt 0[/tex] gives [tex]x\in\left(2k\pi, \frac{\pi}{3}+2k\pi\right)[/tex]. Let [tex]0\lt x\lt\frac{\pi}{3}[/tex] (adding [tex]2k\pi[/tex] changes nothing) and check the triangle inequalities:
[tex]\sin x+\sin 2x-\sin 3x=2\sin x(\cos x-\cos 2x)\gt 0[/tex]
[tex]\sin x+\sin 3x-\sin 2x=\sin 2x(2\cos x-1)\gt 0[/tex]
[tex]\sin 2x+\sin 3x-\sin x=4\sin x\cos\frac{3x}{2}\cos\frac{x}{2}\gt 0[/tex]
The first holds because [tex]0\lt x\lt 2x\lt\pi[/tex], the second because [tex]\cos x\gt\frac{1}{2}[/tex], the third because [tex]\frac{3x}{2}\lt\frac{\pi}{2}[/tex]. So [tex]x\in\left(2k\pi, \frac{\pi}{3}+2k\pi\right),\ k\in\mathbb{Z}[/tex].

Solution:
Let [tex]t=\left|\cos x\right|[/tex], so [tex]0\le t\le 1[/tex] and [tex]\cos 2x=2t^2-1[/tex]. If [tex]t\ge\frac{\sqrt{2}}{2}[/tex], the left side is at least [tex]t\ge\frac{\sqrt{2}}{2}[/tex]. If [tex]0\le t\lt\frac{\sqrt{2}}{2}[/tex], then [tex]2t^2-1\lt 0[/tex] and the left side equals
[tex]f(t)=t+1-2t^2[/tex]
The graph of [tex]f[/tex] is a downward parabola, so on [tex]\left[0, \frac{\sqrt{2}}{2}\right][/tex] its minimum is at an endpoint: [tex]f(0)=1[/tex] and [tex]f\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}[/tex]. So the left side is always at least [tex]\frac{\sqrt{2}}{2}[/tex]. Equality holds for [tex]x=\frac{\pi}{4}+\frac{k\pi}{2}[/tex].

Solution:
The root requires [tex]\pi^2-4x^2\ge 0[/tex], i.e. [tex]-\frac{\pi}{2}\le x\le\frac{\pi}{2}[/tex]. Also [tex]\tan x[/tex] and [tex]\cot x[/tex] must exist, so [tex]x\in\left(-\frac{\pi}{2}, 0\right)\cup\left(0, \frac{\pi}{2}\right)[/tex]. For [tex]x\in\left(-\frac{\pi}{2}, 0\right)[/tex] the left side is negative and the right side is positive, so there are no solutions there. For [tex]x\in\left(0, \frac{\pi}{2}\right)[/tex], by the inequality between the arithmetic and geometric means,
[tex]\tan x+2\cot x\ge 2\sqrt{2\tan x\cot x}=2\sqrt{2}\gt 2[/tex]
The right side is at most [tex]\sqrt[6]{\pi^2}=\sqrt[3]{\pi}\lt\sqrt[3]{8}=2[/tex]. So every [tex]x\in\left(0, \frac{\pi}{2}\right)[/tex] is a solution.

Solution:
Let [tex]t=\sin x-\cos x=\sqrt{2}\sin \left(x-\frac{\pi}{4}\right)[/tex], so [tex]-\sqrt{2}\le t\le\sqrt{2}[/tex]. Since [tex]t^2=1-\sin 2x[/tex], we have [tex]\sin 2x=1-t^2[/tex]. The inequality becomes
[tex]t-t^3\ge\sqrt{2}\iff t^3-t+\sqrt{2}\le 0\iff\left(t+\sqrt{2}\right)\left(t^2-\sqrt{2}t+1\right)\le 0[/tex]
The second factor is positive for all [tex]t[/tex] (negative discriminant), so [tex]t\le-\sqrt{2}[/tex]; together with [tex]t\ge-\sqrt{2}[/tex] this gives [tex]t=-\sqrt{2}[/tex]. Then [tex]\sin \left(x-\frac{\pi}{4}\right)=-1[/tex], so [tex]x-\frac{\pi}{4}=-\frac{\pi}{2}+2k\pi[/tex], i.e. [tex]x=-\frac{\pi}{4}+2k\pi[/tex].

Solution:
Since [tex]x-\frac{\pi}{3}=\left(x+\frac{\pi}{6}\right)-\frac{\pi}{2}[/tex], we have [tex]\sin \left(x-\frac{\pi}{3}\right)=-\cos\left(x+\frac{\pi}{6}\right)[/tex]. With [tex]t=\cos\left(x+\frac{\pi}{6}\right)[/tex]:
[tex]2t^2-3t+1\ge 0\iff(2t-1)(t-1)\ge 0[/tex]
so [tex]t\le\frac{1}{2}[/tex] or [tex]t\ge 1[/tex]. The case [tex]t\ge 1[/tex] means [tex]t=1[/tex]: [tex]x+\frac{\pi}{6}=2k\pi[/tex], [tex]x=-\frac{\pi}{6}+2k\pi[/tex]. The case [tex]t\le\frac{1}{2}[/tex]:
[tex]\frac{\pi}{3}+2k\pi\le x+\frac{\pi}{6}\le\frac{5\pi}{3}+2k\pi\iff\frac{\pi}{6}+2k\pi\le x\le\frac{3\pi}{2}+2k\pi[/tex]
Do not forget the isolated points [tex]x=-\frac{\pi}{6}+2k\pi[/tex].