If A = {x in [0, 2pi] : tan x - tan^2 x 0} | vedclass

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(A) For set $A$: $ an x - an^2 x > 0 \Rightarrow an x(1 - an x) > 0$. This implies $0 < an x < 1$. In the interval $[0, 2\pi]$,this occurs wh

Vedclass · Accepted Answer

$\left(0, \frac{\pi}{6}\right) \cup \left(\pi, \frac{7\pi}{6}\right)$

Vedclass · Answer

(A) For set $A$: $	an x - 	an^2 x > 0 \Rightarrow 	an x(1 - 	an x) > 0$. This implies $0 In the interval $[0, 2\pi]$,this occurs when $x \in (0, \frac{\pi}{4}) \cup (\pi, \frac{5\pi}{4})$. For set $B$: $|\sin x| This implies $x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi]$. Calculating the intersection $A \cap B$: $A = (0, \frac{\pi}{4}) \cup (\pi, \frac{5\pi}{4})$ $B = [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi]$ $A \cap B = (0, \frac{\pi}{6}) \cup (\pi, \frac{7\pi}{6})$.

If $A = \{x \in [0, 2\pi] : \tan x - \tan^2 x > 0\}$ and $B = \{x \in [0, 2\pi] : |\sin x| < \frac{1}{2}\}$,then $A \cap B = $

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