I just want your opinion on my attempt at a solution of this task:

$\displaystyle \tan{\dfrac{x}{2}}>\dfrac{\tan{x}-2}{\tan{x}-2}$

My attempt:

We know that:

$\displaystyle \tan{x}=\dfrac{2\tan{\dfrac{x}{2}}}{1-\tan^2{\dfrac{x}{2}}}$

But, at the beginning we should set limits to tangent function:

$\displaystyle \dfrac{x}{2} \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z} \qquad x \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z} \\ x \neq \pi + 2k\pi,k\in\mathbb{Z} \qquad x \neq \dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}$

If we use the identity that I've given above, we get:

$\displaystyle \tan{\dfrac{x}{2}} > \dfrac{2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2}{2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}}$

Now we can have two cases: denominator is less than zero and denominator is larger than zero.
Let's use substitution $\displaystyle \tan{\dfrac{x}{2}}=m$ for the convenience, then we have quadratic equation:

$\displaystyle -2m^2+2m+2=0$, with the solutions $\displaystyle m_{1,2}=\dfrac{1 \pm \sqrt{5}}{2}$.
Now, denominator is larger than 0 for $\displaystyle m \in \langle \dfrac{1 - \sqrt{5}}{2},\dfrac{1 + \sqrt{5}}{2} \rangle \implies \tan{\dfrac{x}{2}}\in\langle \dfrac{1 - \sqrt{5}}{2},\dfrac{1 + \sqrt{5}}{2} \rangle$.
Now we can multiply inequality with denominator because denominator is not 0 for those values and it is positive so inequality sign won't "change", we have:

$\displaystyle \tan{\dfrac{x}{2}}(2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}) > 2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2$

We have:
$\displaystyle \tan^3{\dfrac{x}{2}} < 1 \implies \tan{\dfrac{x}{2}} < 1$

Then,

$\displaystyle \tan{\dfrac{x}{2}} \in \langle \dfrac{1 - \sqrt{5}}{2},1 \rangle \\ \dfrac{x}{2} \in \langle \arctan{\dfrac{1 - \sqrt{5}}{2}}+k\pi, \dfrac{\pi}{4}+k\pi \rangle ,k\in\mathbb{Z} \\ x \in \langle 2\arctan{\dfrac{1 - \sqrt{5}}{2}}+2k\pi, \dfrac{\pi}{2}+2k\pi \rangle ,k\in\mathbb{Z}$
Which is a good solution because it does not contain our limits for tangent function.

Now for the second case, we assume that: $\displaystyle m \in \langle -\infty,\dfrac{1 - \sqrt{5}}{2} \rangle \cup \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle \implies \tan{\dfrac{x}{2}}\in \langle -\infty,\dfrac{1 - \sqrt{5}}{2} \rangle \cup \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle$

Now we can multiply the inequality with the denominator but the ">" sign will "change":

$\displaystyle \tan{\dfrac{x}{2}}(2+2\tan{\dfrac{x}{2}}-2\tan^2{\dfrac{x}{2}}) < 2\tan^2{\dfrac{x}{2}}+2\tan{\dfrac{x}{2}}-2$

We have:

$\displaystyle \tan^3{\dfrac{x}{2}} > 1 \implies \tan{\dfrac{x}{2}} > 1$

Then,

$\displaystyle \tan{\dfrac{x}{2}} \in \langle \dfrac{1 + \sqrt{5}}{2}, +\infty \rangle \\ \dfrac{x}{2} \in \langle \arctan{\dfrac{1 + \sqrt{5}}{2}}+k\pi, \dfrac{\pi}{2}+k\pi \rangle ,k\in\mathbb{Z} \\ x \in \langle 2\arctan{\dfrac{1 + \sqrt{5}}{2}}+2k\pi, \pi+2k\pi \rangle ,k\in\mathbb{Z}$
Which is a good solution because it does not contain our limits for tangent function.

And finally the solution is:

$\displaystyle x \in \langle 2\arctan{\dfrac{1 - \sqrt{5}}{2}}+2k\pi, \dfrac{\pi}{2}+2k\pi \rangle \cup \langle 2\arctan{\dfrac{1 + \sqrt{5}}{2}}+2k\pi, \pi+2k\pi \rangle ,k\in\mathbb{Z}$