<s>[INST] If $\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$ where $x\in [0,2\pi]$ If we solve the above expression, it’s very clear that the equation doesn’t depend upon x. Then all that is left to is to get 
$$\sin x -\cos x \not =0$$
$$x\not = \pi /4, 5\pi/4$$
But the answer given is $$x\in (0,\pi)-{\pi/4,\pi/2}$$
What is the reason for leaving so many values out ie. $\pi , 2\pi$?
 [/INST] Since
$$\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$$
$$\sin^2x+\sin x\cos x+\cos^2x-\frac{\cos x}{\sqrt{\frac{\sin^2 x+\cos^2x}{\sin^2x}}}-2=-1$$
$$\sin x\cos x-|\sin x|\cos x=0$$
which is always true for $\sin x\ge 0$ otherwise it is true for $x=0,\frac \pi 2, \frac{3\pi}2,2\pi$.
Therefore, since we also need $x\neq \frac \pi 2,\frac {3\pi} 2$ for $\tan x$ and $x\neq  0,2\pi$ for $\cot x$ all the solutions are given by $x\in (0,\pi)$ with $x\neq\pi/4$ and $x\neq \pi/2$.
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