Solve y=a(1+cosθ)^2

$$y = a ( 1 + \cos \theta ) ^ { 2 }$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{y}{\left(\cos(\theta )+1\right)^{2}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\pi \\a\in \mathrm{C}\text{, }&y=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\pi \end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\cos(\theta )\right)^{2}.

$$y=a\left(1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}\right)$$

Use the distributive property to multiply a by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$y=a+2a\cos(\theta )+a\left(\cos(\theta )\right)^{2}$$

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

$$a+2a\cos(\theta )+a\left(\cos(\theta )\right)^{2}=y$$

Combine all terms containing a.

$$\left(1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}\right)a=y$$

The equation is in standard form.

$$\left(\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1\right)a=y$$

Divide both sides by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$\frac{\left(\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1\right)a}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}=\frac{y}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}$$

Dividing by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2} undoes the multiplication by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$a=\frac{y}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}$$

Divide y by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$a=\frac{y}{\left(\cos(\theta )+1\right)^{2}}$$

Solve for a

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\cos(\theta )\right)^{2}.

$$y=a\left(1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}\right)$$

Use the distributive property to multiply a by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$y=a+2a\cos(\theta )+a\left(\cos(\theta )\right)^{2}$$

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

$$a+2a\cos(\theta )+a\left(\cos(\theta )\right)^{2}=y$$

Combine all terms containing a.

$$\left(1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}\right)a=y$$

The equation is in standard form.

$$\left(\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1\right)a=y$$

Divide both sides by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$\frac{\left(\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1\right)a}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}=\frac{y}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}$$

Dividing by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2} undoes the multiplication by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$a=\frac{y}{\left(\cos(\theta )\right)^{2}+2\cos(\theta )+1}$$

Divide y by 1+2\cos(\theta )+\left(\cos(\theta )\right)^{2}.

$$a=\frac{y}{\left(\cos(\theta )+1\right)^{2}}$$

Solve for y

$$y=a\left(\cos(\theta )+1\right)^{2}$$