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. 
 A: Yes.  the equation is equivalent to the couple of systems:
$$
\begin{cases}
\tan x\ge 0\\
\frac{\sin^2 x}{\cos x}\le \frac{3}{2}
\end {cases}
\quad \lor \quad 
\begin{cases}
\tan x< 0\\
-\frac{\sin^2 x}{\cos x}\le \frac{3}{2}
\end {cases}
$$
A: 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 $4ºQ$
$$-\frac{\sin^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-1+\cos^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-2+2\cos^2 x-3\cos x}{2\cos x} \le0$$
A: Let $x\in\left[0,\frac{\pi}{2}\right)\cup\left[\pi,\frac{3\pi}{2}\right)$.
Thus, we need to solve $$\frac{\sin^2x}{\cos{x}}\leq\frac{3}{2}$$ or
$$\frac{(2+\cos x)(2\cos x-1)}{\cos x}\geq0,$$
which is $\cos x\geq\frac{1}{2}$ or $\cos x<0$, which gives $\left[0,\frac{\pi}{3}\right]\cup\left[\pi,\frac{3\pi}{2}\right)$.
Let $x\in\left(\frac{\pi}{2},\pi\right]\cup\left(\frac{3\pi}{2},2\pi\right]$.
Thus, we need to solve $$-\frac{\sin^2x}{\cos{x}}\leq\frac{3}{2}$$ or
$$\frac{(2-\cos x)(2\cos x+1)}{\cos x}\geq0,$$
which is $\cos x\leq-\frac{1}{2}$ or $\cos x>0$, which gives $\left[\frac{2\pi}{3},\pi\right]\cup\left(\frac{3\pi}{2},2\pi\right]$.
