y を解く
\left\{\begin{matrix}y>0\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\beta =2\pi n_{2}+2\pi -\arcsin(\frac{3}{5})\text{ or }\left(\exists n_{4}\in \mathrm{Z}\text{ : }\left(\beta >\frac{\pi \left(4n_{4}+3\right)}{2}\text{ and }\beta <\frac{\pi \left(4n_{4}+5\right)}{2}\right)\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\beta =2\pi n_{3}+\pi -\arcsin(\frac{3}{5})\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\frac{\pi \left(2n_{1}+1\right)}{2}\right)\\y<0\text{, }&\exists n_{5}\in \mathrm{Z}\text{ : }\left(\beta >\frac{\pi \left(4n_{5}+1\right)}{2}\text{ and }\beta <\frac{\pi \left(4n_{5}+3\right)}{2}\right)\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\beta =2\pi n_{3}+\pi -\arcsin(\frac{3}{5})\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.
β を解く
\beta \neq \pi n_{1}+\frac{\pi }{2}
\forall n_{1}\in \mathrm{Z}
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