Solve {l}{tan2alpha=3/4}{tanalpha=n}

$$\left\{ \begin{array} { l } { \tan 2 \alpha = \frac { 3 } { 4 } } \\ { \tan \alpha = n } \end{array} \right.$$

Solve for α, n (complex solution)

$$\alpha =\frac{\pi n_{1}+\arctan(\frac{3}{4})}{2}$$

$$n_{1}\in \mathrm{Z}$$

$$n=\tan(\frac{\pi n_{1}+\arctan(\frac{3}{4})}{2})$$

$$n_{1}\in \mathrm{Z}$$

Solve for α, n

$$\alpha =2\pi n_{2}-\arcsin(\frac{3\sqrt{10}}{10})+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }n=-3\text{, }n_{2}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}-\arcsin(\frac{3\sqrt{10}}{10})+\pi =\frac{\pi \left(2n_{1}+1\right)}{2}$$

$$\alpha =2\pi n_{3}-\arcsin(\frac{3\sqrt{10}}{10})+2\pi \text{, }n_{3}\in \mathrm{Z}\text{, }n=-3\text{, }n_{3}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}-\arcsin(\frac{3\sqrt{10}}{10})+2\pi =\frac{\pi \left(2n_{1}+1\right)}{2}$$

$$\alpha =2\pi n_{4}+\arcsin(\frac{\sqrt{10}}{10})+\pi \text{, }n_{4}\in \mathrm{Z}\text{, }n=\frac{1}{3}\text{, }n_{4}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{4}+\arcsin(\frac{\sqrt{10}}{10})+\pi =\frac{\pi \left(2n_{1}+1\right)}{2}$$

$$\alpha =2\pi n_{5}+\arcsin(\frac{\sqrt{10}}{10})\text{, }n_{5}\in \mathrm{Z}\text{, }n=\frac{1}{3}\text{, }n_{5}\in \mathrm{Z}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{5}+\arcsin(\frac{\sqrt{10}}{10})=\frac{\pi \left(2n_{1}+1\right)}{2}$$