Solve {c}{-1/4m=n*cosθ}{-3/2n=mcosθ}

$$\left. \begin{array} { c } { - \frac { 1 } { 4 } m = n \cdot \cos \theta } \\ { - \frac { 3 } { 2 } n = m \cos \theta } \end{array} \right.$$

Solve for m, n, θ (complex solution)

$$\theta =2\pi n_{1}-\arctan(\frac{\sqrt{15}}{3})+2\pi \text{, }n_{1}\in \mathrm{Z}\text{, }m=-\sqrt{6}n\text{, }n\in \mathrm{C}$$

$$\theta =2\pi n_{2}+\arctan(\frac{\sqrt{15}}{3})\text{, }n_{2}\in \mathrm{Z}\text{, }m=-\sqrt{6}n\text{, }n\in \mathrm{C}$$

$$\theta =2\pi n_{3}+\arctan(\frac{\sqrt{15}}{3})+\pi \text{, }n_{3}\in \mathrm{Z}\text{, }m=\sqrt{6}n\text{, }n\in \mathrm{C}$$

$$\theta =2\pi n_{4}-\arctan(\frac{\sqrt{15}}{3})+\pi \text{, }n_{4}\in \mathrm{Z}\text{, }m=\sqrt{6}n\text{, }n\in \mathrm{C}$$

$$\theta \in \mathrm{C}\text{, }m=0\text{, }n=0$$

Solve for m, n, θ

$$\theta =2\pi n_{1}-\arccos(\frac{\sqrt{6}}{4})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }m=\sqrt{6}n\text{, }n\in \mathrm{R}$$

$$\theta =2\pi n_{2}+\arccos(\frac{\sqrt{6}}{4})+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }m=\sqrt{6}n\text{, }n\in \mathrm{R}$$

$$\theta =2\pi n_{3}+\arccos(\frac{\sqrt{6}}{4})\text{, }n_{3}\in \mathrm{Z}\text{, }m=-\sqrt{6}n\text{, }n\in \mathrm{R}$$

$$\theta =2\pi n_{4}-\arccos(\frac{\sqrt{6}}{4})+2\pi \text{, }n_{4}\in \mathrm{Z}\text{, }m=-\sqrt{6}n\text{, }n\in \mathrm{R}$$

$$\theta \notin \pi n_{5}-\arccos(\frac{\sqrt{6}}{4})+\pi ,\pi n_{5}+\arccos(\frac{\sqrt{6}}{4})\text{, }\forall n_{5}\in \mathrm{Z}\text{, }m=0\text{, }n=0$$