Question Number 74520 by chess1 last updated on 25/Nov/19
Answered by MJS last updated on 26/Nov/19
![0≤x<2π tan x >sin x ((sin x)/(cos x))>sin x case 1: sin x >0 ⇔ 0<x<π (1/(cos x))>1 ⇔ 0<x<(π/2)∨((3π)/2)<x<2π ⇒ 0<x<(π/2) case 2: sin x <0 ⇔ π<x<2π (1/(cos x))<1 ⇔ (π/2)<x<((3π)/2) ⇒ π<x<((3π)/2) ⇒ nπ<x<nπ+(π/2) with n∈Z or x∈ ]nπ; nπ+(π/2)[ with n∈Z](https://www.tinkutara.com/question/Q74564.png)
$$\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$$\mathrm{tan}\:{x}\:>\mathrm{sin}\:{x} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}>\mathrm{sin}\:{x} \\ $$$$\mathrm{case}\:\mathrm{1}:\:\mathrm{sin}\:{x}\:>\mathrm{0}\:\Leftrightarrow\:\mathrm{0}<{x}<\pi \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}}>\mathrm{1}\:\Leftrightarrow\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\vee\frac{\mathrm{3}\pi}{\mathrm{2}}<{x}<\mathrm{2}\pi \\ $$$$\Rightarrow\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{case}\:\mathrm{2}:\:\mathrm{sin}\:{x}\:<\mathrm{0}\:\Leftrightarrow\:\pi<{x}<\mathrm{2}\pi \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}}<\mathrm{1}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}}<{x}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\pi<{x}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:{n}\pi<{x}<{n}\pi+\frac{\pi}{\mathrm{2}}\:\mathrm{with}\:{n}\in\mathbb{Z} \\ $$$$\left.\mathrm{or}\:{x}\in\:\right]{n}\pi;\:{n}\pi+\frac{\pi}{\mathrm{2}}\left[\:\mathrm{with}\:{n}\in\mathbb{Z}\right. \\ $$
Commented by chess1 last updated on 26/Nov/19
$$\mathrm{thanks} \\ $$