Solve 0quadr=a(1-sinθ)

$$0 \quad r = a ( 1 - \sin \theta )$$

Solve for a

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Anything times zero gives zero.

$$0=a\left(1-\sin(\theta )\right)$$

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

$$a\left(1-\sin(\theta )\right)=0$$

Use the distributive property to multiply a by 1-\sin(\theta ).

$$a-a\sin(\theta )=0$$

Combine all terms containing a.

$$\left(1-\sin(\theta )\right)a=0$$

The equation is in standard form.

$$\left(-\sin(\theta )+1\right)a=0$$

Divide 0 by 1-\sin(\theta ).

$$a=0$$

Solve for r

$$r\in \mathrm{R}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }\theta =2\pi n_{1}+\frac{\pi }{2}\text{ or }a=0$$

Steps for Solving Linear Equation

Anything times zero gives zero.

$$0=a\left(1-\sin(\theta )\right)$$

Use the distributive property to multiply a by 1-\sin(\theta ).

$$0=a-a\sin(\theta )$$

The equation is in standard form.

$$0=a\left(-\sin(\theta )+1\right)$$

This is false for any r.

$$r\in$$