Solve quady=Acos(x-B)

$$\quad y = A \cos ( x - B )$$

Solve for A

$$\left\{\begin{matrix}A=\frac{y}{\cos(x-B)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+B+\frac{\pi }{2}\\A\in \mathrm{R}\text{, }&y=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+B+\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$A\cos(x-B)=y$$

The equation is in standard form.

$$\cos(x-B)A=y$$

Divide both sides by \cos(x-B).

$$\frac{\cos(x-B)A}{\cos(x-B)}=\frac{y}{\cos(x-B)}$$

Dividing by \cos(x-B) undoes the multiplication by \cos(x-B).

$$A=\frac{y}{\cos(x-B)}$$