Question

$$\sin ^ { 2 } A - \cos ^ { 2 } A - \cos 2 B = \sin ^ { 2 } B$$

Solve for A (complex solution)

$A=\left(-i\right)\ln(\left(-\frac{1}{2}\right)\left(\left(-1\right)e^{\left(-2i\right)B}+\left(-1\right)e^{2iB}-2+\left(-1\right)\left(\left(e^{2iB}+e^{\left(-2i\right)B}+6\right)\left(e^{2iB}+e^{\left(-2i\right)B}-2\right)\right)^{\frac{1}{2}}\right)^{\frac{1}{2}})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$
$A=i\ln(2)+\left(-\frac{1}{2}i\right)\ln(\left(-1\right)e^{\left(-2i\right)B}+\left(-1\right)e^{2iB}-2+\left(-1\right)\left(\left(e^{2iB}+e^{\left(-2i\right)B}+6\right)\left(e^{2iB}+e^{\left(-2i\right)B}-2\right)\right)^{\frac{1}{2}})+2n_{3}\pi \text{, }n_{3}\in \mathrm{Z}$
$A=i\ln(2)+\left(-\frac{1}{2}i\right)\ln(\left(-1\right)e^{\left(-2i\right)B}+\left(-1\right)e^{2iB}-2+\left(\left(e^{2iB}+e^{\left(-2i\right)B}+6\right)\left(e^{2iB}+e^{\left(-2i\right)B}-2\right)\right)^{\frac{1}{2}})+2n_{33}\pi \text{, }n_{33}\in \mathrm{Z}$
$A=\left(-i\right)\ln(\left(-\frac{1}{2}\right)\left(\left(-1\right)e^{\left(-2i\right)B}+\left(-1\right)e^{2iB}-2+\left(\left(e^{2iB}+e^{\left(-2i\right)B}+6\right)\left(e^{2iB}+e^{\left(-2i\right)B}-2\right)\right)^{\frac{1}{2}}\right)^{\frac{1}{2}})+2\pi n_{52}\text{, }n_{52}\in \mathrm{Z}$

Solve for A

$A=\arcsin(\frac{\sqrt{\cos(2B)+3}}{2})+\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$
$A=-\arcsin(\frac{\sqrt{\cos(2B)+3}}{2})+\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$