<s>[INST] Determine the set of all values of $ x \in [0, 2 \pi] $ 
Determine the set of all values of $ x \in [0, 2 \pi] $ that simultaneously satisfy $ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1} < 0 $ and $  \tan x + \sqrt {3} < (1+ \sqrt {3} \cot x) \cot x $

My ''solution'':
$\frac{2{{\sin }^{2}}x+\sin x-1}{\cos x-1}<0\Rightarrow 2{{\sin }^{2}}x+\sin x-1>0\Rightarrow \sin x\in \left( \frac{1}{2},1 \right]$ and $\tan x+\sqrt{3}<(1+\sqrt{3}\cot x)\cot x\Leftrightarrow \tan x+\sqrt{3}<\frac{\tan x+\sqrt{3}}{\tan x}\cot x\Leftrightarrow \tan x+\sqrt{3}<\left( \tan x+\sqrt{3} \right)\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$

Am I right? Is there any better approach?
 [/INST] The first inequality,
$$ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1}=\frac{(2\sin x -1)(\sin x+1)}{ \cos x-1}< 0 $$
Then,
$\sin x+1>0,\>\>\>\cos x-1<0 \implies \sin x > \frac12\implies x\in (\frac \pi6,\frac{5\pi}6)\tag{1}$
The second inequality,
$$\tan x+\sqrt{3}<\left( \tan x+\sqrt{3} \right)\cot^2x$$
Case 1. 
$\tan x+\sqrt{3}>0,\>\>\>\cot^2x>1 \implies 
x\in (0,\frac\pi4),\>(\frac{3\pi}4,\frac{5\pi}4),\>(\frac{7\pi}4,2\pi)\tag{2}$ 
Case 2.
$\tan x+\sqrt{3}<0,\>\>\>\cot^2x<1 \implies 
x\in (\frac{\pi}2,\frac{2\pi}3),\>(\frac{3\pi}2,\frac{5\pi}3)\tag{3}$ 
Combine (2) and (3),
$x \in (0,\frac\pi4),(\frac\pi2,\frac{2\pi}3),(\frac{3\pi}4,\frac{5\pi}4),(\frac{3\pi}2,\frac{5\pi}3),(\frac{7\pi}4,2\pi)\tag{4}$
Thus, from (1) and (4), determine the set of all values 
$$x \in (\frac\pi6,\frac\pi4),(\frac\pi2,\frac{2\pi}3),(\frac{3\pi}4,\frac{5\pi}6)$$
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