Solve {l}{0=2tanθ-20(1/acosθ)^2}{-1=5/2tanθ-31,25(1/acosθ)^2}

$$\left\{ \begin{array} { l } { 0 = 2 \tan \theta - 20 ( \frac { 1 } { a \cos \theta } ) ^ { 2 } } \\ { - 1 = \frac { 5 } { 2 } \tan \theta - 31,25 ( \frac { 1 } { a \cos \theta } ) ^ { 2 } } \end{array} \right.$$

Solve for θ, a

$$\theta =2\pi n_{1}+\arcsin(\frac{8\sqrt{89}}{89})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }a=-\frac{\sqrt{89}}{2}\approx -4.716990566$$

$$\theta =2\pi n_{1}+\arcsin(\frac{8\sqrt{89}}{89})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }a=\frac{\sqrt{89}}{2}\approx 4.716990566$$

$$\theta =2\pi n_{2}+\arcsin(\frac{8\sqrt{89}}{89})\text{, }n_{2}\in \mathrm{Z}\text{, }a=-\frac{\sqrt{89}}{2}\approx -4.716990566$$

$$\theta =2\pi n_{2}+\arcsin(\frac{8\sqrt{89}}{89})\text{, }n_{2}\in \mathrm{Z}\text{, }a=\frac{\sqrt{89}}{2}\approx 4.716990566$$