y=ln(sin1/x) шешу

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

y мәнін табыңыз (complex solution)

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

$$x\neq 0\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }x=\frac{1}{2\pi n_{2}}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{i}{2\pi n_{1}i+\pi i}\text{ and }\nexists n_{3}\in \mathrm{Z}\text{ : }x=\frac{1}{\pi n_{3}}$$

y мәнін табыңыз

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

$$x>\frac{1}{\pi }\text{ or }\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{2}}\text{ and }x<0\right)$$

$$not(n_{2}>-1)\text{ and }not(n_{2}=0)\text{ or }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{1}+\pi }\text{ and }x<\frac{1}{2\pi n_{1}}\text{ and }n_{1}\neq 0\right)\text{, }not(n_{1}=0)\text{ and }not(n_{1}<1)\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{2}+\pi }\text{ and }x<\frac{1}{2\pi n_{2}}\text{ and }n_{2}\neq 0\right)\text{, }not(n_{2}=0)\text{ and }not(n_{2}<1)\right)\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{1}}\text{ and }x<0\right)\text{, }not(n_{1}>-1)\text{ and }not(n_{1}=0)\text{ or }\left(x<0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{1}}\text{ and }n_{1}\neq 0\right)\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\left(x>\frac{1}{2\pi n_{2}}\text{ and }n_{2}\neq 0\right)\right)$$