Решить {left({e}^frac{1{x}}-1right)}^2{left(tan(1/x)right)}^3>0

Решение для x

\left\{\begin{matrix}x\in \mathrm{R}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{2}}\text{ and }x<0\right)\text{, }\text{false}\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\left(x>\frac{2}{4\pi n_{3}+\pi }\text{ and }x<-\frac{2}{\pi -4\pi n_{3}}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x\neq 0\\x\in \mathrm{R}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{2}+\pi }\text{ and }x<\frac{1}{2\pi n_{2}}\right)\text{, }not(n_{2}=0)\text{ and }not(n_{2}<1)\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\left(x>\frac{2}{4\pi n_{3}+\pi }\text{ and }x<-\frac{2}{\pi -4\pi n_{3}}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x\neq 0\\x\in \mathrm{R}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x>\frac{2}{\pi }\\x\in \mathrm{R}\text{, }&\exists n_{4}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{4}}\text{ and }x<-\frac{1}{\pi -2\pi n_{4}}\right)\text{ and }\exists n_{5}\in \mathrm{Z}\text{ : }\left(x>-\frac{2}{\pi -4\pi n_{5}}\text{ and }x<-\frac{2}{3\pi -4\pi n_{5}}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x\neq 0\\x\in \mathrm{R}\text{, }&\exists n_{4}\in \mathrm{Z}\text{ : }\left(x>0\text{ and }x<\frac{1}{2\pi n_{4}}\right)\text{, }\text{false}\text{ and }\exists n_{5}\in \mathrm{Z}\text{ : }\left(x>-\frac{2}{\pi -4\pi n_{5}}\text{ and }x<-\frac{2}{3\pi -4\pi n_{5}}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x\neq 0\\x\in \mathrm{R}\text{, }&\exists n_{5}\in \mathrm{Z}\text{ : }\left(x>-\frac{2}{\pi -4\pi n_{5}}\text{ and }x<-\frac{2}{3\pi -4\pi n_{5}}\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }\text{ and }x<-\frac{1}{\pi }\end{matrix}\right,