y=ln(1-sinx/1+sinx) шешу

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

y мәнін табыңыз (complex solution)

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

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

x мәнін табыңыз

$$x=\left(-1\right)arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})+2\pi n_{28}\text{, }n_{28}\in \mathrm{Z}\text{, }\exists n_{9}\in \mathrm{Z}\text{ : }\left(n_{9}<\left(\left(-1\right)arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})+2\pi n_{28}+\left(-\frac{1}{2}\right)\pi \right)\pi ^{-1}\text{ and }\left(-1\right)arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})+2\pi n_{28}<\pi \left(n_{9}+\frac{3}{2}\right)\right)\text{ and }\exists n_{9}\in \mathrm{Z}\text{ : }\left(\left(-1\right)arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})+2\pi n_{28}>\pi n_{9}+\frac{1}{2}\pi \text{ and }\left(-1\right)arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})+2\pi n_{28}<\pi \left(n_{9}+\frac{3}{2}\right)\right)$$

$$x=\pi +2\pi n_{23}+arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})\text{, }n_{23}\in \mathrm{Z}\text{, }\exists n_{9}\in \mathrm{Z}\text{ : }\left(\pi +2\pi n_{23}+arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})>\pi n_{9}+\frac{1}{2}\pi \text{ and }\pi +2\pi n_{23}+arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})<\left(n_{9}+\frac{3}{2}\right)\pi \right)\text{ and }\exists n_{9}\in \mathrm{Z}\text{ : }\left(\pi +2\pi n_{23}+arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})>\pi n_{9}+\frac{1}{2}\pi \text{ and }\pi +2\pi n_{23}+arcSin(\left(-1+e^{y}\right)\left(1+e^{y}\right)^{-1})<\pi \left(n_{9}+\frac{3}{2}\right)\right)$$

y мәнін табыңыз

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

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(x>\pi n_{1}+\frac{\pi }{2}\text{ and }x<\pi n_{1}+\frac{3\pi }{2}\right)$$