Solve $2\tan{2x}\leq3\tan{x}.$ Problem:
$$\text{Solve} \quad 2\tan{2x}\leq3\tan{x}.$$
A problem of this character will yield 5 points on an exam. However, having the correct answer does not suffice to get all the 5 points. Full stringency and mathematical accuracy, on top of a correct final answer, warrants full house. I've decided to present a (partial) solution here and I want you to help me really to comb through it and search for possible logical loopholes.

Solution attempt:
Application of the double angle formula for $\tan{x}$ on LHS yields:
$$\text{LHS}=2\cdot\frac{\sin{2x}}{\cos{2x}}=2\cdot\frac{2\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}}=[\text{Divide by} \ \cos^2{x}\neq0]=\frac{4\tan{x}}{1-\tan^2{x}}.$$
Setting $t=\tan{x}$ and moving the RHS over and subtracting gives the equivalent inequality:
$$\frac{4t}{1-t^2}-3t=\frac{4t-3t(1-t^2)}{1-t^2}=\frac{t(1+3t^2)}{(1+t)(1-t)}=p(t)\leq0.$$
Since the factor $(1+3t^2)>0, \ \forall x\in \mathbb{R}, $ it suffices to examine the signs of the factors $t, \ (1+t), \ (1-t)$ and the entire expression that I denoted $p(t).$ The following table emerges:
\begin{array}
{|l|cr}
t=  & -\infty & -1 & \ & 0 & \ & 1 & +\infty\\
\hline
1+t & - & 0 & + &  & + &  & +\\
\hline
t   & - &   & - & 0 & + &  & +\\
\hline
1-t & + &   & + &  & + & 0 & -\\
\hline
p(t)& + & \varnothing & - & 0 & + & \varnothing & -\\
\end{array}
This indicates that the solutionset of $p(t)\leq0$ is given by $t\in(-1,0]\cup(1,\infty).$ Hereafter I'm stuck, I don't know how to revert to $x$. How do I do this in an effective way?
 A: The goal is to find conditions on $x$ such that $\tan(x)$ will be in the indicated intervals.  Recall that if $x\in \left[ -\frac{\pi}{2},\frac{\pi}{2}\right] =: \mathcal{D}$, then
$$\tan(x) = y \iff \arctan(y) = x.$$
It isn't too difficult to see that
$$ \arctan(-1) = -\frac{\pi}{4}, \qquad\text{and}\qquad \arctan(0) = 0.$$
Since $\arctan$ is increasing and continuous on its domain, it follows that $\tan(x) \in (-1,0]$ and $x\in\mathcal{D}$ if and only if
$$ x \in (\arctan(-1),\arctan(0)] = \left( -\frac{\pi}{4}, 0\right].$$
By a similar argument, we conclude that $\tan(x) \in (1,\infty)$ and $x\in\mathcal{D}$ if and only if
$$ x \in \left(\arctan(1),\lim_{y\to\infty} \arctan(y)\right) = \left( \frac{\pi}{4}, \frac{\pi}{2} \right).$$
As the tangent function is periodic with fundamental period equal to $\pi$, it follows that if $x$ satisfies the given inequality, then so to will $x + k\pi$ for any $k \in\mathbb{Z}$.  Therefore the complete set of solutions is given by
$$ \bigcup_{k\in\mathbb{Z}} \left[ \left( -\frac{\pi}{4} + k\pi, k\pi\right] \cup \left( \frac{\pi}{4} + k\pi, \frac{\pi}{2}+k\pi \right) \right].$$
A: Duplication formula for tangent
$\dfrac{4 \tan  x}{1-\tan ^2 x}\leq 3 \tan  x$
Bring all in the LHS
$\dfrac{4 \tan  x}{1-\tan ^2 x}-3 \tan  x\leq 0$
add together
$-\dfrac{\tan  x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\leq 0$
which is better as
$\dfrac{\tan  x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\geq 0$
The parenthesis $\left(3 \tan ^2 x+1\right)$ is positive for any $x$ because is the sum of two squares so we need to see where
$\tan x \geq 0$ verified in $0\leq x <\dfrac{\pi}{2}\lor \pi\leq x <\dfrac{3\pi}{2}$
and $\tan ^2 x-1>0$ verified when $\tan x<-1\lor \tan x>1$
which is $\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{\pi }{2}<x<\dfrac{3 \pi }{4}\lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{3 \pi }{2}<x<\dfrac{7 \pi }{4}$

The solution is then
$\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{3 \pi }{4}<x\leq\pi \lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{7 \pi }{4}<x\leq2 \pi$
A: We need to solve
$$\frac{4\tan{x}}{1-\tan^2x}\leq3\tan{x}$$ or
$$\tan{x}\left(\frac{4}{1-\tan^2x}-3\right)\leq0$$ or
$$\frac{\tan{x}(3\tan^2x+1)}{(1-\tan{x})(1+\tan{x})}\leq0,$$
which by the intervals method gives
$$\tan{x}<-1$$ or $$0\leq\tan{x}<1$$ and we got the answer:
$$\left[\pi k\leq x<\frac{\pi}{4}+\pi k\right)\cup\left(-\frac{\pi}{2}+\pi k,-\frac{\pi}{4}+\pi k\right),$$
where $k\in\mathbb Z$.
For example, the inequality $\tan{x}<-1$ we can solve by the following way.
Draw the trigonometric circle (the circle $x^2+y^2=1$) and the tangent to the circle in the point $B(1,0)$.
This line names a tangents-axis.
Let $A(1,a)$ placed in the tangents-axis.
Thus, easy to see that $\tan\measuredangle AOB=a$ and since we need to solve $\tan{x}<-1$,
we got $-\frac{\pi}{2}<x<-\frac{\pi}{4}$ (see on the tangents-axis).
Since the period of $\tan$ is equal to $\pi$, finally we obtain:
$$-\frac{\pi}{2}+\pi k<x<-\frac{\pi}{4}+\pi k,$$
where $k\in\mathbb Z$.
By the similar way we can solve $0\leq\tan{x}<1$.
