Solve in [0;2π] 2sin(4x−(π/6))≥1 Please sirs...


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Question Number 94910 by mathocean1 last updated on 21/May/20

$Solve in [0; 2 π]$ $2 \sin (4 x - \frac{π}{6}) ⩾ 1$ $Please sirs . . .$
	Answered by mr W last updated on 21/May/20

	$\sin (4 x - \frac{π}{6}) ⩾ \frac{1}{2}$ $2 k π + \frac{π}{6} ⩽ 4 x - \frac{π}{6} ⩽ 2 k π + π - \frac{π}{6}$ $\Rightarrow \frac{k π}{2} + \frac{π}{12} ⩽ x ⩽ \frac{k π}{2} + \frac{π}{4} w i t h k \in Z$ $w i t h i n [0, 2 π] :$ $\frac{π}{12} ⩽ x ⩽ \frac{π}{4}$ $\frac{7 π}{12} ⩽ x ⩽ \frac{3 π}{4}$ $\frac{13 π}{12} ⩽ x ⩽ \frac{5 π}{4}$ $\frac{9 π}{12} ⩽ x ⩽ \frac{7 π}{4}$
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