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Question Number 77472 by mathocean1 last updated on 06/Jan/20

$$\mathrm{Hello}\mathrm{sirs}...$$$$\mathrm{i}\mathrm{want}\mathrm{that}\mathrm{you}\mathrm{explain}\mathrm{me}\mathrm{how}$$$$\mathrm{we}\mathrm{solve}\mathrm{the}\mathrm{trigonometric}$$$$\mathrm{inequality}\mathrm{with}\mathrm{tangente}.$$$$\mathrm{we}\mathrm{can}\mathrm{take}\mathrm{for}\mathrm{example}\tan 2\mathrm{x}⩾\sqrt{3}$$$$\mathrm{i}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{type}\mathrm{of}\mathrm{equality}$$$$\mathrm{but}\mathrm{not}\mathrm{the}\mathrm{inequality}!\mathrm{Please}$$$$\mathrm{i}\mathrm{need}\mathrm{your}\mathrm{help}...\mathrm{Even}\mathrm{the}\mathrm{steps}$$$$\mathrm{to}\mathrm{make}\mathrm{graphic}\mathrm{to}\mathrm{read}\mathrm{the}\mathrm{solu}-$$$$\mathrm{tion}.$$

Commented by MJS last updated on 06/Jan/20

$$\left(1\right)\mathrm{solve}\mathrm{the}\mathrm{equality}\mathrm{for}x\mathrm{within}\mathrm{one}\mathrm{period}$$$$\tan 2x=\sqrt{3};-\frac{\pi }{4}⩽x<\frac{\pi }{4}$$$$\Rightarrow x=\frac{\pi }{6}$$$$\left(2\right)\mathrm{calculate}\mathrm{the}\mathrm{gradient}\mathrm{of}\mathrm{the}\mathrm{curve}$$$$\frac{d}{dx}\left[\tan 2x\right]=\frac{2}{{\cos }^{2}2x}>0\mathrm{\forall }x\in \mathbb{R}\mathrm{\setminus }\{-\frac{\pi }{4}+\frac{\pi }{2}n\};n\in \mathbb{Z}$$$$\mathrm{also}\tan 2x\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x=-\frac{\pi }{4}+\frac{\pi }{2}n$$$$\Rightarrow \mathrm{within}\mathrm{this}\mathrm{period}\tan 2x⩾\sqrt{3}\mathrm{with}\frac{\pi }{6}⩽x<\frac{\pi }{4}$$$$\Rightarrow \tan 2x⩾\sqrt{3}\mathrm{with}\frac{\pi }{6}+\frac{\pi }{2}n⩽x<\frac{\pi }{4}+\frac{\pi }{2}n;n\in \mathbb{Z}$$

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

$$\mathrm{the}\mathrm{gradient}\mathrm{in}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{is}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{derivate}$$$$P=\left(\begin{array}{c}p\\ f\left(p\right)\end{array}\right)\Rightarrow \mathrm{gradient}=f{}^{\prime }\left(p\right)$$$$f{}^{\prime }\left(p\right)>0\Rightarrow \mathrm{curve}\mathrm{is}\mathrm{increasing}\mathrm{in}P$$$$f{}^{\prime }\left(p\right)=0\Rightarrow \mathrm{curve}\mathrm{is}\mathrm{flat}\mathrm{in}P$$$$f{}^{\prime }\left(p\right)<0\Rightarrow \mathrm{curve}\mathrm{is}\mathrm{decreasing}\mathrm{in}P$$

Commented by mathocean1 last updated on 06/Jan/20

$$\mathrm{Thank}\mathrm{you}...$$

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