Solve $2\tan{2x}\leq3\tan{x}.$

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 $$ \bigcup_{k\in\mathbb{Z}} \left[ \left( -\frac{\pi}{4} + k\pi, k\pi\right] \cup \left( \frac{\pi}{4} + k\pi, \frac{\pi}{2}+k\pi \right) \right].$$