<s>[INST] Solve the inequality $\sin(x)\cdot|\tan{x}|\le\frac{3}{2}$ As in the title, solve the inequality $\sin(x)\cdot|\tan{x}|\le\frac{3}{2}$ for $x\in[0;2\pi]$. My concern is that I don't knwo how to get rid of the absolute value sign - should I consider separately cases of the $\tan{x}$ being positive and negative? 
Any hints greatly appreciated. 
 [/INST] Hint:
\begin{cases}
\tan x &, 1º Q &\text{or}&3ºQ\\
-\tan x &, 2º Q &\text{or}&4ºQ
\end{cases}
Now solve in each case.
$1.$ $1º Q$ or $3ºQ$
$$\frac{\sin^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{1-\cos^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{2-2\cos^2 x-3\cos x}{2\cos x} \le0$$
$1.$ $2º Q$ or $4ºQ$
$$-\frac{\sin^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-1+\cos^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-2+2\cos^2 x-3\cos x}{2\cos x} \le0$$
</s>