Selesaikan untuk p
p=\tan(\theta )
\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}
Selesaikan untuk 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)
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Had
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