Rešite v=tane^x

$$v = \tan e ^ { x }$$

Rešitev za v

$$v=\tan(e^{x})$$

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

Rešitev za x

$$\left\{\begin{matrix}x=\ln(2\pi n_{2}+\arcsin(\frac{v}{\sqrt{v^{2}+1}})+\pi )\text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{4}\in \mathrm{Z}\text{ : }\left(n_{4}\text{bmod}2=0\text{ and }n_{2}\geq \frac{n_{4}}{2}-\frac{1}{2}\text{ and }n_{2}\leq \frac{n_{4}}{2}+\frac{1}{2}\right)\text{ and }n_{2}\geq 0\text{, }&n_{1}\geq 0\\x=\ln(2\pi n_{3}+\arcsin(\frac{v}{\sqrt{v^{2}+1}}))\text{, }n_{3}\in \mathrm{Z}\text{, }\left(n_{3}\geq 1\text{ and }\exists n_{4}\in \mathrm{Z}\text{ : }\left(n_{4}\text{bmod}2=1\text{ and }n_{3}\geq \frac{n_{4}}{2}\text{ and }n_{3}\leq \frac{n_{4}}{2}+1\right)\right)\text{ or }\left(v>0\text{ and }n_{3}\geq 0\text{ and }\exists n_{4}\in \mathrm{Z}\text{ : }\left(n_{4}\text{bmod}2=1\text{ and }n_{3}\geq \frac{n_{4}}{2}\text{ and }n_{3}\leq \frac{n_{4}}{2}+1\right)\right)\text{, }&n_{2}\geq 0\text{ and }n_{1}\geq 0\end{matrix}\right,$$