Hello-solve-in-0-2pi-tan2x-3-

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 ?