求解 {l}{sin3/5cosbeta=-5/13}{tan(a-beta)=22}

$$\left. \begin{array}{l}{ \sin \frac { 3 } { 5 } \cos \beta = - \frac { 5 } { 13 } }\\{ \tan ( a - \beta ) = 22 }\end{array} \right.$$

求解 β, a 的值

$$\left\{\begin{matrix}\beta =2\pi n_{1}-i\ln(\frac{\sqrt{24395964603148870561481014759128391}+156250000000000000}{4254162297224853})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }a\in \cup n_{3},2\pi n_{1}+2\pi n_{3}-i\ln(\frac{\sqrt{24395964603148870561481014759128391}+156250000000000000}{4254162297224853})+\pi +\arctan(22)\text{, }\nexists n_{4}\in \mathrm{Z}\text{ : }n_{3}=\frac{-6n_{4}-\frac{2\arctan(22)}{\pi }+3}{4}\text{, }n_{1}\in \mathrm{Z}\text{; }\beta =2\pi n_{2}-i\ln(\frac{156250000000000000-\sqrt{24395964603148870561481014759128391}}{4254162297224853})+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }a\in \cup n_{3},2\pi n_{2}+2\pi n_{3}-i\ln(\frac{156250000000000000-\sqrt{24395964603148870561481014759128391}}{4254162297224853})+\pi +\arctan(22)\text{, }\nexists n_{4}\in \mathrm{Z}\text{ : }n_{3}=\frac{-6n_{4}-\frac{2\arctan(22)}{\pi }+3}{4}\text{, }n_{2}\in \mathrm{Z}\text{, }&\exists n_{3}\in \mathrm{Z}\text{ : }\left(\nexists n_{4}\in \mathrm{Z}\text{ : }n_{3}=\frac{-6n_{4}-\frac{2\arctan(22)}{\pi }+3}{4}\right)\\\beta =2\pi n_{1}-i\ln(\frac{\sqrt{24395964603148870561481014759128391}+156250000000000000}{4254162297224853})+\pi \text{, }n_{1}\in \mathrm{Z}\text{, }a\in \cup n_{5},2\pi n_{1}+2\pi n_{5}-i\ln(\frac{\sqrt{24395964603148870561481014759128391}+156250000000000000}{4254162297224853})+\arctan(22)+2\pi \text{, }\nexists n_{4}\in \mathrm{Z}\text{ : }n_{4}=\frac{-4n_{5}-\frac{2\arctan(22)}{\pi }+1}{6}\text{, }n_{1}\in \mathrm{Z}\text{; }\beta =2\pi n_{2}-i\ln(\frac{156250000000000000-\sqrt{24395964603148870561481014759128391}}{4254162297224853})+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }a\in \cup n_{5},2\pi n_{2}+2\pi n_{5}-i\ln(\frac{156250000000000000-\sqrt{24395964603148870561481014759128391}}{4254162297224853})+\arctan(22)+2\pi \text{, }\nexists n_{4}\in \mathrm{Z}\text{ : }n_{4}=\frac{-4n_{5}-\frac{2\arctan(22)}{\pi }+1}{6}\text{, }n_{2}\in \mathrm{Z}\text{, }&\exists n_{5}\in \mathrm{Z}\text{ : }\left(\nexists n_{4}\in \mathrm{Z}\text{ : }n_{4}=\frac{-4n_{5}-\frac{2\arctan(22)}{\pi }+1}{6}\right)\end{matrix}\right.$$