ដោះស្រាយ {l}{tanalpha=2}{1/sinalpha=m}

$$\left. \begin{array} { l } { \tan \alpha = 2 } \\ { \frac { 1 } { \sin \alpha } = m } \end{array} \right.$$

ដោះស្រាយសម្រាប់ α, m (complex solution)

$$\alpha =2\pi n_{1}+\arctan(2)\text{, }n_{1}\in \mathrm{Z}\text{, }m=\frac{\sqrt{5}}{2}\approx 1.118033989$$

$$\alpha =2\pi n_{2}+\pi +\arctan(2)\text{, }n_{2}\in \mathrm{Z}\text{, }m=-\frac{\sqrt{5}}{2}\approx -1.118033989$$

ដោះស្រាយសម្រាប់ α, m

$$\alpha =2\pi n_{2}+\arcsin(\frac{2\sqrt{5}}{5})+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }m=-\frac{\sqrt{5}}{2}\text{, }n_{2}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arcsin(\frac{2\sqrt{5}}{5})+\pi =\pi n_{1}$$

$$\alpha =2\pi n_{3}+\arcsin(\frac{2\sqrt{5}}{5})\text{, }n_{3}\in \mathrm{Z}\text{, }m=\frac{\sqrt{5}}{2}\text{, }n_{3}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+\arcsin(\frac{2\sqrt{5}}{5})=\pi n_{1}$$