Solve sin2alpha=1/9quadtext{if}sinalpha

$$\sin 2 \alpha = \frac { 1 } { 9 } \quad \text { if } \sin \alpha$$

Solve for f

$$\left\{\begin{matrix}f=-18i\cos(\alpha )\text{, }&\nexists n_{2}\in \mathrm{Z}\text{ : }\alpha =\pi n_{2}\\f\in \mathrm{C}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$\frac{1}{9}if\sin(\alpha )=\sin(2\alpha )$$

The equation is in standard form.

$$\frac{i\sin(\alpha )}{9}f=\sin(2\alpha )$$

Divide both sides by \frac{1}{9}i\sin(\alpha ).

$$\frac{9\times \frac{i\sin(\alpha )}{9}f}{i\sin(\alpha )}=\frac{9\sin(2\alpha )}{i\sin(\alpha )}$$

Dividing by \frac{1}{9}i\sin(\alpha ) undoes the multiplication by \frac{1}{9}i\sin(\alpha ).

$$f=\frac{9\sin(2\alpha )}{i\sin(\alpha )}$$

Divide \sin(2\alpha ) by \frac{1}{9}i\sin(\alpha ).

$$f=-18i\cos(\alpha )$$