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Question Number 74520 by chess1 last updated on 25/Nov/19

Answered by MJS last updated on 26/Nov/19

$$0⩽x<2\pi$$$$\tan x>\sin x$$$$\frac{\sin x}{\cos x}>\sin x$$$$\mathrm{case}1:\sin x>0\iff 0<x<\pi$$$$\frac{1}{\cos x}>1\iff 0<x<\frac{\pi }{2}\vee \frac{3\pi }{2}<x<2\pi$$$$\Rightarrow 0<x<\frac{\pi }{2}$$$$\mathrm{case}2:\sin x<0\iff \pi <x<2\pi$$$$\frac{1}{\cos x}<1\iff \frac{\pi }{2}<x<\frac{3\pi }{2}$$$$\Rightarrow \pi <x<\frac{3\pi }{2}$$$$$$$$\Rightarrow n\pi <x<n\pi +\frac{\pi }{2}\mathrm{with}n\in \mathbb{Z}$$$$\mathrm{or}x\in ]n\pi ;n\pi +\frac{\pi }{2}[\mathrm{with}n\in \mathbb{Z}$$

Commented by chess1 last updated on 26/Nov/19

$$\mathrm{thanks}$$

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