Question Number 194766 by dimentri last updated on 15/Jul/23
$$\:\:\:\mathrm{1}+\mathrm{2cot}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\:\mathrm{3}\: \\ $$$$\:\:\:{x}=? \\ $$
Answered by Frix last updated on 15/Jul/23
$$\mathrm{1}+\mathrm{2cor}\:\mathrm{2}{x}\:\mathrm{cot}\:{x}\:=\mathrm{3} \\ $$$$\mathrm{cot}^{\mathrm{2}} \:{x}\:=\mathrm{3} \\ $$$$\mathrm{cot}\:{x}\:=\pm\sqrt{\mathrm{3}} \\ $$$${x}=\pm\frac{\pi}{\mathrm{6}}+{n}\pi;\:{n}\in\mathbb{Z} \\ $$
Commented by dimentri last updated on 15/Jul/23
$$\:\: \\ $$
Answered by MM42 last updated on 15/Jul/23
$${tan}\mathrm{2}{x}×{tanx}−\mathrm{3}{tan}\mathrm{2}{xtanx}+\mathrm{2}=\mathrm{0} \\ $$$${let}\:\:\:{tanx}={u} \\ $$$$\frac{\mathrm{2}{u}^{\mathrm{2}} }{\mathrm{1}−{u}^{\mathrm{2}} }−\frac{\mathrm{6}{u}^{\mathrm{2}} }{\mathrm{1}−{u}^{\mathrm{2}} }+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{u}={tanx}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\Rightarrow{x}={k}\pi\pm\frac{\pi}{\mathrm{6}}\:\checkmark \\ $$$$ \\ $$
Commented by dimentri last updated on 15/Jul/23
$$\:\gtrdot \\ $$