ແກ້ +tanθ=p

$$+ \tan \theta = p$$

ແກ້ສຳລັບ p

$$p=\tan(\theta )$$

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

ແກ້ສຳລັບ θ

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

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