Solve {l}{R(sin2beta-cos2beta)=7/2}{R(-1+cos2beta+sin2beta)=2}

$$\left\{ \begin{array} { l } { R ( \sin 2 \beta - \cos 2 \beta ) = \frac { 7 } { 2 } } \\ { R ( - 1 + \cos 2 \beta + \sin 2 \beta ) = 2 } \end{array} \right.$$

Solve for R, β (complex solution)

$$R=\frac{13}{2}=6.5\text{, }\beta =\frac{2\pi n_{1}+\arctan(\frac{12}{5})}{2}\text{, }n_{1}\in \mathrm{Z}$$

$$R=-\frac{5}{2}=-2.5\text{, }\beta =\frac{2\pi n_{2}+2\pi -\arctan(\frac{3}{4})}{2}\text{, }n_{2}\in \mathrm{Z}$$

Solve for R, β

$$\left\{\begin{matrix}R=\frac{13}{2}=6.5\text{, }\beta \in \cup n_{2},\frac{2\pi n_{2}+\arcsin(\frac{12}{13})}{2}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arcsin(\frac{12}{13})=\frac{\pi \left(4n_{1}+1\right)}{4}\text{, }n_{2}\in \mathrm{Z}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arcsin(\frac{12}{13})=\frac{\pi \left(4n_{1}+1\right)}{4}\right)\right)\\R=-\frac{5}{2}=-2.5\text{, }\beta \in \cup n_{3},\frac{2\pi n_{3}+2\pi -\arcsin(\frac{3}{5})}{2}\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+2\pi -\arcsin(\frac{3}{5})=\frac{\pi \left(4n_{1}+1\right)}{4}\text{, }n_{3}\in \mathrm{Z}\text{, }&\exists n_{3}\in \mathrm{Z}\text{ : }\left(\exists n_{3}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+2\pi -\arcsin(\frac{3}{5})=\frac{\pi \left(4n_{1}+1\right)}{4}\right)\right)\end{matrix}\right.$$