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Question Number 76110 by mathocean1 last updated on 23/Dec/19

$$hello\mathrm{solve}\mathrm{in}\mathbb{R}$$ $$\tan x>\sqrt{3}$$ $$pleaseexplainmeifpossible.$$

Answered by MJS last updated on 24/Dec/19

$$\tan x=\sqrt{3}\wedge 0⩽x<2\pi$$ $$\Rightarrow x=\frac{\pi }{3}\vee x=\frac{4\pi }{3}$$ $$\mathrm{but}y=\tan x\mathrm{changes}\mathrm{its}\mathrm{sign}\mathrm{at}x=0,x=\frac{\pi }{2},$$ $$x=\pi \mathrm{and}x=\frac{3\pi }{2}$$ $$\Rightarrow \tan x>\sqrt{3}\iff \frac{\pi }{3}<x<\frac{\pi }{2}\vee \frac{4\pi }{3}<x<\frac{3\pi }{2}$$ $$\mathrm{now}\mathrm{if}x\in \mathbb{R}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{in}\mathrm{each}\mathrm{period}\mathrm{of}\mathrm{length}\pi$$ $$\Rightarrow \tan x>\sqrt{3}\iff \frac{\pi }{3}+n\pi <x<\frac{\pi }{2}+n\pi \mathrm{with}n\in \mathbb{Z}$$

Commented bybenjo last updated on 24/Dec/19

$$\mathrm{great}\mathrm{sir}...$$

Commented bymathocean1 last updated on 24/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}...$$

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