y=operatorname{ctg}x を解きます| Microsoft 数学ソルバー

$$y = \operatorname { ctg } x$$

x を解く

$$x=2n_{3}\pi +\left(-1\right)\pi +ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})\text{, }n_{3}\in \mathrm{Z}\text{, }\exists n_{362}\in \mathrm{Z}\text{ : }\left(n_{3}>\left(-\frac{1}{2}\right)\left(\left(-1\right)\pi +ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})+\left(-1\right)\pi n_{362}\right)\pi ^{-1}\text{ and }2n_{3}\pi +\left(-1\right)\pi +ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})<\pi \left(n_{362}+1\right)\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }2n_{3}\pi +\left(-1\right)\pi +ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})=\pi n_{1}$$

$$x=ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})+2\pi n_{341}\text{, }n_{341}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})+2\pi n_{341}=\pi n_{1}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }ArcCosI(y\left(1+y^{2}\right)^{-\frac{1}{2}})+2\pi n_{341}=\pi n_{1}$$

y を解く

$$y=\cot(x)$$

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