Solving Trigonometric Inequalities: Difficult Problems with Solutions

Solution:
On the unit circle the inequality gives
[tex]x\in\left(-\frac{\pi}{6}+2k\pi, \frac{\pi}{3}+2k\pi\right]\cup\left[\frac{2\pi}{3}+2k\pi, \frac{7\pi}{6}+2k\pi\right)[/tex]
With [tex]\pi\approx 3. 1416[/tex], the parts inside [tex]\left[0, 10\right][/tex] are approximately [tex]\left[0, 1. 05\right][/tex], [tex]\left[2. 09, 3. 67\right)[/tex], [tex]\left(5. 76, 7. 33\right][/tex] and [tex]\left[8. 38, 9. 95\right)[/tex]. They contain the integers [tex]0, 1, 3, 6, 7, 9[/tex], so there are [tex]6[/tex] integer solutions.

Solution:
Domain: [tex]\cos x\ne-1[/tex] and [tex]\sin x\ne-1[/tex]; then both denominators are positive. Multiplying by [tex](1+\cos x)(1+\sin x)\gt 0[/tex] and using [tex]\sin ^2 x+\cos^2 x=1[/tex]:
[tex]\sin x+\sin ^2 x+\cos x+\cos^2 x\ge 1+\sin x+\cos x+\sin x\cos x\iff\sin x\cos x\le 0[/tex]
So [tex]x\in\left[-\frac{\pi}{2}+2k\pi, 2k\pi\right]\cup\left[\frac{\pi}{2}+2k\pi, \pi+2k\pi\right][/tex]; the points [tex]x=-\frac{\pi}{2}+2k\pi[/tex] and [tex]x=\pi+2k\pi[/tex] are not in the domain and must be removed.

Solution:
If [tex]x\le 0[/tex], then [tex]\sin |x|=\sin (-x)=-\sin x[/tex] and the left side is [tex]0[/tex], so every [tex]x\le 0[/tex] is a solution. If [tex]x\gt 0[/tex], the inequality is [tex]2\sin x\le 1[/tex], i.e. [tex]\sin x\le\frac{1}{2}[/tex]:
[tex]x\in\left[\frac{5\pi}{6}+2k\pi, \frac{13\pi}{6}+2k\pi\right][/tex]
For [tex]x\gt 0[/tex] this gives [tex]\left(0, \frac{\pi}{6}\right][/tex] and the intervals with [tex]k=0, 1, 2, \ldots[/tex] Together with [tex]x\le 0[/tex] we get the answer.

Solution:
Move everything to the left and group the terms:
[tex]2\sin x(2\cos x+1)-\sqrt{2}(2\cos x+1)\lt 0\iff(2\cos x+1)\left(2\sin x-\sqrt{2}\right)\lt 0[/tex]
Case 1: [tex]\cos x\gt-\frac{1}{2}[/tex] and [tex]\sin x\lt\frac{\sqrt{2}}{2}[/tex], which gives [tex]x\in\left(-\frac{2\pi}{3}+2k\pi, \frac{\pi}{4}+2k\pi\right)[/tex]. Case 2: [tex]\cos x\lt-\frac{1}{2}[/tex] and [tex]\sin x\gt\frac{\sqrt{2}}{2}[/tex], which gives [tex]x\in\left(\frac{2\pi}{3}+2k\pi, \frac{3\pi}{4}+2k\pi\right)[/tex]. The answer is the union of the two cases.

Solution:
Since [tex]\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}[/tex], we can write
[tex]4\sin x+6\cos x=2\sqrt{13}\sin (x+\varphi), \quad\cos\varphi=\frac{2}{\sqrt{13}}, \ \sin \varphi=\frac{3}{\sqrt{13}}, \ \varphi=\arctan \frac{3}{2}[/tex]
The inequality becomes [tex]\left|\sin (x+\varphi)\right|\lt 1[/tex]. It fails only when [tex]\sin (x+\varphi)=\pm 1[/tex], i.e. [tex]x+\varphi=\frac{\pi}{2}+k\pi[/tex]. So [tex]x\ne\frac{\pi}{2}-\arctan \frac{3}{2}+k\pi=\arctan \frac{2}{3}+k\pi[/tex].

Solution:
Using [tex]\sin 3x=3\sin x-4\sin ^3 x[/tex]:
[tex]2\sin ^2 x-\sin x+\sin 3x=2\sin x+2\sin ^2 x-4\sin ^3 x=2\sin x(1-\sin x)(1+2\sin x)[/tex]
The factor [tex]1-\sin x[/tex] is never negative and equals [tex]0[/tex] only when [tex]\sin x=1[/tex]. So we need [tex]\sin x(1+2\sin x)\gt 0[/tex] and [tex]\sin x\ne 1[/tex], i.e. [tex]0\lt\sin x\lt 1[/tex] or [tex]\sin x\lt-\frac{1}{2}[/tex]. The first gives [tex]x\in\left(2k\pi, \pi+2k\pi\right)[/tex] without [tex]x=\frac{\pi}{2}+2k\pi[/tex]; the second gives [tex]x\in\left(\frac{7\pi}{6}+2k\pi, \frac{11\pi}{6}+2k\pi\right)[/tex].

Solution:
Let [tex]t=\cos x[/tex]. The root exists for [tex]4t+2\ge 0[/tex], i.e. [tex]t\ge-\frac{1}{2}[/tex]. If [tex]3t+\frac{1}{2}\lt 0[/tex], i.e. [tex]-\frac{1}{2}\le t\lt-\frac{1}{6}[/tex], the inequality is true. If [tex]t\ge-\frac{1}{6}[/tex], we square both sides:
[tex]4t+2\ge 9t^2+3t+\frac{1}{4}\iff 36t^2-4t-7\le 0\iff-\frac{7}{18}\le t\le\frac{1}{2}[/tex]
Together with [tex]t\ge-\frac{1}{6}[/tex] this gives [tex]-\frac{1}{6}\le t\le\frac{1}{2}[/tex]. Joining the two cases: [tex]-\frac{1}{2}\le\cos x\le\frac{1}{2}[/tex], so [tex]x\in\left[\frac{\pi}{3}+k\pi, \frac{2\pi}{3}+k\pi\right],\ k\in\mathbb{Z}[/tex].