y=1/ln(2sinx+1) を解きます| Microsoft 数学ソルバー

$$y = \frac { 1 } { \ln ( 2 \sin x + 1 ) }$$

x を解く (複素数の解)

$$\left\{\begin{matrix}x=-i\ln(\frac{ie^{\frac{1}{y}}}{2}-\frac{i\sqrt{e^{\frac{2}{y}}-2e^{\frac{1}{y}}-3}}{2}-\frac{1}{2}i)+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&y\neq 0\text{ and }\frac{-ie^{\frac{1}{y}}+i\sqrt{e^{\frac{2}{y}}-2e^{\frac{1}{y}}-3}}{2}\neq -\frac{1}{2}i\text{ and }Im(\ln(e^{\frac{1}{y}}))-Im(\frac{1}{y})=0\\x=-i\ln(\frac{ie^{\frac{1}{y}}}{2}+\frac{i\sqrt{e^{\frac{2}{y}}-2e^{\frac{1}{y}}-3}}{2}-\frac{1}{2}i)+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&y\neq 0\text{ and }\frac{-ie^{\frac{1}{y}}-i\sqrt{e^{\frac{2}{y}}-2e^{\frac{1}{y}}-3}}{2}\neq -\frac{1}{2}i\text{ and }Im(\ln(e^{\frac{1}{y}}))-Im(\frac{1}{y})=0\end{matrix}\right.$$

y を解く (複素数の解)

$$y=\frac{1}{\ln(2\sin(x)+1)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }\left(x=2\pi n_{1}+\frac{11\pi }{6}\text{ or }x=2\pi n_{1}+\frac{7\pi }{6}\right)\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}$$

y を解く

$$y=\frac{1}{\ln(2\sin(x)+1)}$$

$$\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>2\pi n_{1}+\frac{11\pi }{6}\text{ and }x<2\pi n_{1}+\frac{19\pi }{6}\right)$$