פתור את y=tan(2x-pi)

$$y= \tan ( 2x- \pi )$$

פתור עבור x

$$x=\frac{2\pi n_{1}+\arcsin(\frac{y}{\sqrt{y^{2}+1}})+2\pi }{2}\text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-1}{4}\text{ and }n_{1}<\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }+1}{4}\right)$$

$$x=\frac{2\pi n_{2}+\arcsin(\frac{y}{\sqrt{y^{2}+1}})+\pi }{2}\text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}>\frac{4n_{2}+\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-3}{2}\text{ and }n_{3}<\frac{4n_{2}+\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-1}{2}\right)$$

פתור עבור y

$$y=\tan(2x)$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}$$