Question Number 77472 by mathocean1 last updated on 06/Jan/20
$$\mathrm{Hello}\:\mathrm{sirs}…\: \\ $$$$\mathrm{i}\:\mathrm{want}\:\mathrm{that}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\: \\ $$$$\:\mathrm{we}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{trigonometric} \\ $$$$\mathrm{inequality}\:\mathrm{with}\:\mathrm{tangente}.\: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{take}\:\mathrm{for}\:\mathrm{example}\:\mathrm{tan2x}\geqslant\sqrt{\mathrm{3}} \\ $$$$\mathrm{i}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{type}\:\mathrm{of}\:\mathrm{equality} \\ $$$$\mathrm{but}\:\mathrm{not}\:\mathrm{the}\:\mathrm{inequality}!\:\mathrm{Please}\: \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}…\:\mathrm{Even}\:\mathrm{the}\:\mathrm{steps} \\ $$$$\mathrm{to}\:\mathrm{make}\:\mathrm{graphic}\:\mathrm{to}\:\mathrm{read}\:\mathrm{the}\:\mathrm{solu}− \\ $$$$\mathrm{tion}. \\ $$
Commented by MJS last updated on 06/Jan/20
![(1) solve the equality for x within one period tan 2x =(√3); −(π/4)≤x<(π/4) ⇒ x=(π/6) (2) calculate the gradient of the curve (d/dx)[tan 2x]=(2/(cos^2 2x))>0∀x∈R\{−(π/4)+(π/2)n}; n∈Z also tan 2x is not defined for x=−(π/4)+(π/2)n ⇒ within this period tan 2x ≥(√3) with (π/6)≤x<(π/4) ⇒ tan 2x ≥(√3) with (π/6)+(π/2)n≤x<(π/4)+(π/2)n; n∈Z](https://www.tinkutara.com/question/Q77476.png)
$$\left(\mathrm{1}\right)\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equality}\:\mathrm{for}\:{x}\:\mathrm{within}\:\mathrm{one}\:\mathrm{period} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\sqrt{\mathrm{3}};\:−\frac{\pi}{\mathrm{4}}\leqslant{x}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}=\frac{\pi}{\mathrm{6}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\:\:\:\:\:\frac{{d}}{{dx}}\left[\mathrm{tan}\:\mathrm{2}{x}\right]=\frac{\mathrm{2}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}>\mathrm{0}\forall{x}\in\mathbb{R}\backslash\left\{−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}{n}\right\};\:{n}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\mathrm{also}\:\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}=−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}{n} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{within}\:\mathrm{this}\:\mathrm{period}\:\mathrm{tan}\:\mathrm{2}{x}\:\geqslant\sqrt{\mathrm{3}}\:\mathrm{with}\:\frac{\pi}{\mathrm{6}}\leqslant{x}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{tan}\:\mathrm{2}{x}\:\geqslant\sqrt{\mathrm{3}}\:\mathrm{with}\:\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{2}}{n}\leqslant{x}<\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}{n};\:{n}\in\mathbb{Z} \\ $$
Commented by mathocean1 last updated on 06/Jan/20
$$\mathrm{Thank}\:\mathrm{very}\:\mathrm{much}… \\ $$
Commented by mathocean1 last updated on 06/Jan/20
$$\mathrm{I}\:\mathrm{haven}''\mathrm{t}\:\mathrm{studied}\:\mathrm{about}\:\mathrm{gradient} \\ $$$$\mathrm{of}\:\mathrm{curve}\:\mathrm{sir}…\:\mathrm{what}\:\mathrm{is}\:\mathrm{it}? \\ $$
Commented by MJS last updated on 06/Jan/20
$$\mathrm{the}\:\mathrm{gradient}\:\mathrm{in}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{derivate} \\ $$$${P}=\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix}\:\Rightarrow\:\mathrm{gradient}\:=\:{f}\:'\left({p}\right) \\ $$$${f}\:'\left({p}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{curve}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{in}\:{P} \\ $$$${f}\:'\left({p}\right)=\mathrm{0}\:\Rightarrow\:\mathrm{curve}\:\mathrm{is}\:\mathrm{flat}\:\mathrm{in}\:{P} \\ $$$${f}\:'\left({p}\right)<\mathrm{0}\:\Rightarrow\:\mathrm{curve}\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{in}\:{P} \\ $$
Commented by mathocean1 last updated on 06/Jan/20
$$\mathrm{Thank}\:\mathrm{you}… \\ $$