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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.

3 Answers 3

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Yes. the equation is equivalent to the couple of systems: $$ \begin{cases} \tan x\ge 0\\ \frac{\sin^2 x}{\cos x}\le \frac{3}{2} \end {cases} \quad \lor \quad \begin{cases} \tan x< 0\\ -\frac{\sin^2 x}{\cos x}\le \frac{3}{2} \end {cases} $$

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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 $4ºQ$

$$-\frac{\sin^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-1+\cos^2 x}{\cos x} \le\frac{3}{2}\rightarrow \frac{-2+2\cos^2 x-3\cos x}{2\cos x} \le0$$

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    This is kind of an odd way to tell the OP "yes, your idea to consider each case separately is good" – 2017-01-03
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    Well, guess I'm too tired, this was pretty obvious... Thanks anyway +1 – 2017-01-03
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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 $\left[\frac{2\pi}{3},\pi\right]\cup\left(\frac{3\pi}{2},2\pi\right]$.