Question Number 79075 by mathocean1 last updated on 22/Jan/20
![Hello solve in [0;2π] tan2x≥(√3)](Q79075.png)
$$\mathrm{Hello}\: \\ $$$$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\right]\:\mathrm{tan2}{x}\geqslant\sqrt{\mathrm{3}} \\ $$
Answered by mr W last updated on 22/Jan/20
![tan 2x≥(√3) ⇒kπ+(π/3)≤2x<kπ+(π/2) ⇒((kπ)/2)+(π/6)≤x<((kπ)/2)+(π/(4 )) such that x is in [0, 2π] k=0,1,2,3 ⇒(π/6)≤x<(π/(4 )) ⇒(π/2)+(π/6)=((2π)/3)≤x<(π/2)+(π/(4 ))=((3π)/4) ⇒π+(π/6)=((7π)/6)≤x<π+(π/(4 ))=((5π)/4) ⇒((3π)/2)+(π/6)=((5π)/3)≤x<((3π)/2)+(π/(4 ))=((7π)/4)](Q79076.png)
$$\mathrm{tan}\:\mathrm{2}{x}\geqslant\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{k}\pi+\frac{\pi}{\mathrm{3}}\leqslant\mathrm{2}{x}<{k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\leqslant{x}<\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}\:} \\ $$$${such}\:{that}\:{x}\:{is}\:{in}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{6}}\leqslant{x}<\frac{\pi}{\mathrm{4}\:} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}=\frac{\mathrm{2}\pi}{\mathrm{3}}\leqslant{x}<\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}\:}=\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\Rightarrow\pi+\frac{\pi}{\mathrm{6}}=\frac{\mathrm{7}\pi}{\mathrm{6}}\leqslant{x}<\pi+\frac{\pi}{\mathrm{4}\:}=\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{3}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}=\frac{\mathrm{5}\pi}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{3}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}\:}=\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$
Commented by mathocean1 last updated on 22/Jan/20
$$\mathrm{please}\:\mathrm{sir}\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{determinate}\:\mathrm{the}\: \\ $$$$\mathrm{final}\:\mathrm{interval}\:\mathrm{of}\:{x}\:? \\ $$