Hello solve in [0;2π] tan2x≥(√3)


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Question Number 79075 by mathocean1 last updated on 22/Jan/20

$Hello$ $solve in [0; 2 π] \tan 2 x ⩾ \sqrt{3}$
	Answered by mr W last updated on 22/Jan/20

	$\tan 2 x ⩾ \sqrt{3}$ $\Rightarrow k π + \frac{π}{3} ⩽ 2 x < k π + \frac{π}{2}$ $\Rightarrow \frac{k π}{2} + \frac{π}{6} ⩽ x < \frac{k π}{2} + \frac{π}{4}$ $s u c h t h a t x i s i n [0, 2 π] k = 0, 1, 2, 3$ $\Rightarrow \frac{π}{6} ⩽ x < \frac{π}{4}$ $\Rightarrow \frac{π}{2} + \frac{π}{6} = \frac{2 π}{3} ⩽ x < \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4}$ $\Rightarrow π + \frac{π}{6} = \frac{7 π}{6} ⩽ x < π + \frac{π}{4} = \frac{5 π}{4}$ $\Rightarrow \frac{3 π}{2} + \frac{π}{6} = \frac{5 π}{3} ⩽ x < \frac{3 π}{2} + \frac{π}{4} = \frac{7 π}{4}$
	Commented by mathocean1 last updated on 22/Jan/20

	$please sir how can i determinate the$ $final interval of x ?$
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