Question Number 76110 by mathocean1 last updated on 23/Dec/19
$${hello}\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\mathrm{tan}{x}>\sqrt{\mathrm{3}} \\ $$$${please}\:{explain}\:{me}\:{if}\:{possible}. \\ $$
Answered by MJS last updated on 24/Dec/19
$$\mathrm{tan}\:{x}\:=\sqrt{\mathrm{3}}\wedge\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$$\Rightarrow\:{x}=\frac{\pi}{\mathrm{3}}\vee{x}=\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$$$\mathrm{but}\:{y}=\mathrm{tan}\:{x}\:\mathrm{changes}\:\mathrm{its}\:\mathrm{sign}\:\mathrm{at}\:{x}=\mathrm{0},\:{x}=\frac{\pi}{\mathrm{2}}, \\ $$$${x}=\pi\:\mathrm{and}\:{x}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{tan}\:{x}\:>\sqrt{\mathrm{3}}\:\Leftrightarrow\:\frac{\pi}{\mathrm{3}}<{x}<\frac{\pi}{\mathrm{2}}\:\vee\:\frac{\mathrm{4}\pi}{\mathrm{3}}<{x}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{if}\:{x}\in\mathbb{R}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{each}\:\mathrm{period}\:\mathrm{of}\:\mathrm{length}\:\pi \\ $$$$\Rightarrow\:\mathrm{tan}\:{x}\:>\sqrt{\mathrm{3}}\:\Leftrightarrow\:\frac{\pi}{\mathrm{3}}+{n}\pi<{x}<\frac{\pi}{\mathrm{2}}+{n}\pi\:\mathrm{with}\:{n}\in\mathbb{Z} \\ $$
Commented by benjo last updated on 24/Dec/19
$$\mathrm{great}\:\mathrm{sir}... \\ $$
Commented by mathocean1 last updated on 24/Dec/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}... \\ $$