Solve for A (complex solution)
\left\{\begin{matrix}A=\frac{2ye^{ix-iB}}{e^{2ix-2iB}+1}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=3\pi n_{1}+B-\frac{3\pi }{2}\\A\in \mathrm{C}\text{, }&y=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=3\pi n_{1}+B-\frac{3\pi }{2}\end{matrix}\right.
Solve for A
\left\{\begin{matrix}A=\frac{y}{\cos(x-B)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+B+\frac{\pi }{2}\\A\in \mathrm{R}\text{, }&y=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+B+\frac{\pi }{2}\end{matrix}\right.
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A\cos(x-B)=y
Swap sides so that all variable terms are on the left hand side.
\cos(x-B)A=y
The equation is in standard form.
\frac{\cos(x-B)A}{\cos(x-B)}=\frac{y}{\cos(x-B)}
Divide both sides by \cos(x-B).
A=\frac{y}{\cos(x-B)}
Dividing by \cos(x-B) undoes the multiplication by \cos(x-B).
A\cos(x-B)=y
Swap sides so that all variable terms are on the left hand side.
\cos(x-B)A=y
The equation is in standard form.
\frac{\cos(x-B)A}{\cos(x-B)}=\frac{y}{\cos(x-B)}
Divide both sides by \cos(x-B).
A=\frac{y}{\cos(x-B)}
Dividing by \cos(x-B) undoes the multiplication by \cos(x-B).
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