求解 y=tanx

$$y = \tan x$$

求解 x 的值

$$x=\pi +2n_{3}\pi +arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\text{, }n_{3}\in \mathrm{Z}\text{, }\exists n_{42}\in \mathrm{Z}\text{ : }\left(n_{3}>\left(-\frac{1}{2}\right)\left(\frac{1}{2}\pi +\left(-1\right)\pi n_{42}+arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\right)\pi ^{-1}\text{ and }n_{3}<\left(-\frac{1}{2}\right)\left(\left(-\frac{1}{2}\right)\pi +\left(-1\right)\pi n_{42}+arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\right)\pi ^{-1}\right)$$

$$x=arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{22}\text{, }n_{22}\in \mathrm{Z}\text{, }\exists n_{42}\in \mathrm{Z}\text{ : }\left(n_{42}<\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{22}+\left(-\frac{1}{2}\right)\pi \right)\pi ^{-1}\text{ and }n_{42}>\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{22}+\left(-\frac{3}{2}\right)\pi \right)\pi ^{-1}\right)$$

求解 y 的值

$$y=\tan(x)$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$$