Question Number 78040 by Pratah last updated on 13/Jan/20
Commented by mr W last updated on 13/Jan/20
$${the}\:{same}\:{path}\:{as}\:{for} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$
Commented by MJS last updated on 14/Jan/20
$$\mathrm{Cardano}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{since}\:\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{Trigonometric}\:\mathrm{solution} \\ $$
Commented by jagoll last updated on 14/Jan/20
$${how}\:{by}\:{letting}\:\mathrm{2}{x}\:=\:\mathrm{sin}\:{t}\:? \\ $$$$\mathrm{sin}\:^{\mathrm{3}} {t}−\mathrm{3sin}\:{t}+\mathrm{1}=\mathrm{0} \\ $$
Commented by mr W last updated on 14/Jan/20
$${you}\:{may}\:{not}\:{let}\:\mathrm{2}{x}=\mathrm{sin}\:{t}.\:{this}\:{way} \\ $$$${you}\:{assume}\:−\frac{\mathrm{1}}{\mathrm{2}}<{x}<\frac{\mathrm{1}}{\mathrm{2}}\:! \\ $$
Answered by jagoll last updated on 13/Jan/20
$${using}\:{Cardano}\:{Method} \\ $$