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 For set $B$: $|\sin x| To find $A \cap B$,we look for the intersection of these intervals:$A \cap B = \left( \left(0, \frac{\pi}{4}ight) \cup \left(\pi, \frac{5\pi}{4}ight) ight) \cap \left( \left[0, \frac{\pi}{6}ight) \cup \left(\frac{5\pi}{6}, \frac{7\pi}{6}ight) \cup \left(\frac{11\pi}{6}, 2\piight] ight)$.Intersection: $\left(0, \frac{\pi}{6}ight) \cup \left(\pi, \frac{7\pi}{6}ight)$.

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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