Solve for A
A=\frac{\cos(2a)}{2}
Solve for a
a=\arccos(\frac{\sqrt{2\left(2A+1\right)}}{2})+\pi n_{1}-\pi \text{, }n_{1}\in \mathrm{Z}
a=-\arccos(\frac{\sqrt{2\left(2A+1\right)}}{2})+\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }|A|\leq \frac{1}{2}
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2A=\cos(2a)
The equation is in standard form.
\frac{2A}{2}=\frac{\cos(2a)}{2}
Divide both sides by 2.
A=\frac{\cos(2a)}{2}
Dividing by 2 undoes the multiplication by 2.
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