Question Number 192343 by cortano12 last updated on 15/May/23
$$\:\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{y}=\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$
Answered by manxsol last updated on 15/May/23
$$\:\:{x}\neq{k}\pi\frac{\pi}{\mathrm{2}}\:\:\:\:{y}>\mathrm{0} \\ $$$$\mathrm{2}{ysin}^{\mathrm{2}} {x}=\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} \\ $$$${sin}^{\mathrm{4}} {x}−\left(\mathrm{2}+\mathrm{2}{y}\right){sin}^{\mathrm{2}} {x}+\mathrm{1} \\ $$$$\Delta>\mathrm{0} \\ $$$$\left(\mathrm{2}+\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{4}>\mathrm{0} \\ $$$${y}<−\mathrm{2}\:\:\:\cap\:\:\:{y}>\mathrm{0} \\ $$$${sol}\:\:{y}>\mathrm{0}−\left\{{f}\left({x}\right)/{x}={k}\pi/\mathrm{2}\right\} \\ $$$${min}\:{y}>\mathrm{0} \\ $$$$ \\ $$
Commented by mehdee42 last updated on 15/May/23
$${ok} \\ $$$${there}\:{is}\:{not}\:{minimum} \\ $$