Solve y=Acosx+B

$$y = A \cos x + B$$

Solve for A (complex solution)

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

Steps for Solving Linear Equation

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

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

Subtract B from both sides.

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

The equation is in standard form.

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

Divide both sides by \cos(x).

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

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

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

Solve for A

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

Steps for Solving Linear Equation

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

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

Subtract B from both sides.

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

The equation is in standard form.

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

Divide both sides by \cos(x).

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

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

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

Solve for B

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