Solving Trigonometric Inequalities: Problems with Solutions

Solution:
The function [tex]\tan x[/tex] has period [tex]\pi[/tex] and increases on [tex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex]. Since [tex]\tan 0=0[/tex] and [tex]\tan \frac{\pi}{3}=\sqrt{3}[/tex], on this interval
[tex]0\lt\tan x\lt\sqrt{3}\iff 0\lt x\lt\frac{\pi}{3}[/tex]
Adding the period: [tex]x\in\left(k\pi, \frac{\pi}{3}+k\pi\right),\ k\in\mathbb{Z}[/tex].

Solution:
Since [tex]\sin x\cos x=\frac{1}{2}\sin 2x[/tex], the inequality becomes
[tex]\sin 2x\gt 1[/tex]
But [tex]\sin 2x\le 1[/tex] for every [tex]x[/tex], so the inequality has no solutions.

Solution:
Let [tex]t=3x-\frac{\pi}{4}[/tex]. Then [tex]\sin t\gt\frac{\sqrt{3}}{2}[/tex], which on the unit circle means
[tex]\frac{\pi}{3}+2k\pi\lt t\lt\frac{2\pi}{3}+2k\pi\iff\frac{7\pi}{12}+2k\pi\lt 3x\lt\frac{11\pi}{12}+2k\pi[/tex]
Dividing by [tex]3[/tex]: [tex]\frac{7\pi}{36}+\frac{2k\pi}{3}\lt x\lt\frac{11\pi}{36}+\frac{2k\pi}{3},\ k\in\mathbb{Z}[/tex].

Solution:
Factor:
[tex]\left(\sin x-\frac{1}{2}\right)\left(\sin x+\frac{1}{2}\right)\le 0\iff-\frac{1}{2}\le\sin x\le\frac{1}{2}[/tex]
In [tex]\left(0, 2\pi\right)[/tex] the condition [tex]\sin x\le\frac{1}{2}[/tex] gives [tex]\left(0, \frac{\pi}{6}\right]\cup\left[\frac{5\pi}{6}, 2\pi\right)[/tex], and [tex]\sin x\ge-\frac{1}{2}[/tex] gives [tex]\left(0, \frac{7\pi}{6}\right]\cup\left[\frac{11\pi}{6}, 2\pi\right)[/tex]. The answer is the intersection of the two sets.

Solution:
Since [tex]\cot \left(\frac{\pi}{2}-\alpha\right)=\tan \alpha[/tex], we have [tex]\cot \left(\frac{\pi}{4}-x\right)=\tan \left(\frac{\pi}{4}+x\right)[/tex], and
[tex]\tan t\ge\sqrt{3}\iff\frac{\pi}{3}+k\pi\le t\lt\frac{\pi}{2}+k\pi[/tex]
With [tex]t=\frac{\pi}{4}+x[/tex] we get [tex]\frac{\pi}{12}+k\pi\le x\lt\frac{\pi}{4}+k\pi[/tex]. Only [tex]k=-1[/tex] gives points in [tex]\left[-\pi, 0\right][/tex]: [tex]-\frac{11\pi}{12}\le x\lt-\frac{3\pi}{4}[/tex]. The smallest solution is [tex]x=-\frac{11\pi}{12}[/tex].