Question

$$\cos A = \frac { 3 } { 5 } , \frac { \sin A - \cos A } { 2 \tan A }$$

Solve for A, a (complex solution)

$A=2\pi n_{1}+2\pi -\arctan(\frac{4}{3})\text{, }n_{1}\in \mathrm{Z}\text{, }a=\frac{21}{40}=0.525$
$A=2\pi n_{2}+\arctan(\frac{4}{3})\text{, }n_{2}\in \mathrm{Z}\text{, }a=\frac{3}{40}=0.075$

Solve for A, a

$\left\{\begin{matrix}A\in \cup n_{2},2\pi n_{2}+\arccos(\frac{3}{5})\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arccos(\frac{3}{5})=\pi n_{1}\text{, }n_{2}\in \mathrm{Z}\text{, }a=\frac{3}{40}=0.075\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\left(\exists n_{2}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{2}+\arccos(\frac{3}{5})=\pi n_{1}\right)\right)\\A\in \cup n_{3},2\pi n_{3}+2\pi -\arccos(\frac{3}{5})\text{, }\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+2\pi -\arccos(\frac{3}{5})=\pi n_{1}\text{, }n_{3}\in \mathrm{Z}\text{, }a=\frac{21}{40}=0.525\text{, }&\exists n_{3}\in \mathrm{Z}\text{ : }\left(\exists n_{3}\in \mathrm{Z}\text{ : }\left(\nexists n_{1}\in \mathrm{Z}\text{ : }2\pi n_{3}+2\pi -\arccos(\frac{3}{5})=\pi n_{1}\right)\right)\end{matrix}\right.$