Solve {l}{A+B=90}{sin(A-B)=1/2}{a=A}{b=B}{text{Solvefor}c,dtext{where}}{c=a}{d=b}

$$\left. \begin{array} { l } { A + B = 90 }\\ { \sin(A - B) = \frac{1}{2} }\\ { a = A }\\ { b = B }\\ { \text{Solve for } c,d \text{ where} } \\ { c = a }\\ { d = b } \end{array} \right.$$

Solve for A, B, a, b, c, d (complex solution)

$$A=90-B$$

$$B\in \cup n_{1},\pi n_{1}-\frac{169\pi }{12}+45$$

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

$$a=90-B$$

$$b\in \cup n_{1},\pi n_{1}-\frac{169\pi }{12}+45$$

$$B=\pi n_{1}-\frac{169\pi }{12}+45$$

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

$$c\in \cup n_{1},\cup n_{1},\cup n_{1},\pi n_{1}-\frac{167\pi }{12}+45$$

$$B=-\pi n_{1}+\frac{167\pi }{12}+45$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }b=\pi n_{1}-\frac{169\pi }{12}+45\text{, }B=n_{1}\pi +45+\left(-\frac{169}{12}\right)\pi \right)\text{, }n_{1}\in \mathrm{Z}\text{, }d\in \cup n_{1},\cup n_{1},\cup n_{1},\cup n_{1},\pi n_{1}-\frac{169\pi }{12}+45\text{, }b=\pi n_{1}-\frac{169\pi }{12}+45\text{ and }B=\pi n_{1}-\frac{169\pi }{12}+45\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }B=\pi n_{1}-\frac{169\pi }{12}+45\right)\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }c=\pi n_{1}-\frac{167\pi }{12}+45\text{, }B=45+\left(-1\right)n_{1}\pi +\frac{167}{12}\pi \right)\right)\text{, }\exists n_{1}\in \mathrm{Z}\text{ : }\left(\exists n_{1}\in \mathrm{Z}\text{ : }b=n_{1}\pi +45+\left(-\frac{169}{12}\right)\pi \text{, }B=n_{1}\pi +45+\left(-\frac{169}{12}\right)\pi \right)\right)\text{, }n_{1}\in \mathrm{Z}$$

Solve for A, B, a, b, c, d

$$A=\pi n_{1}-\frac{167\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}\text{, }B=\pi n_{1}-\frac{169\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}\text{, }a=\pi n_{1}-\frac{167\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}\text{, }b=\pi n_{1}-\frac{169\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}\text{, }c=\pi n_{1}-\frac{167\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}\text{, }d=\pi n_{1}-\frac{169\pi }{12}+45\text{, }n_{1}\in \mathrm{Z}$$

$$A=\pi n_{2}-\frac{163\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}\text{, }B=\pi n_{2}-\frac{161\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}\text{, }a=\pi n_{2}-\frac{163\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}\text{, }b=\pi n_{2}-\frac{161\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}\text{, }c=\pi n_{2}-\frac{163\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}\text{, }d=\pi n_{2}-\frac{161\pi }{12}+45\text{, }n_{2}\in \mathrm{Z}$$