Solve 5sqrt{2}text{y}sinalpha=frac{2sqrt{5}}{5}

$$5 \sqrt { 2 } \text { y } \sin \alpha = \frac { 2 \sqrt { 5 } } { 5 }$$

Solve for y

$$y=\frac{\sqrt{10}}{25\sin(\alpha )}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 5.

$$25\sqrt{2}y\sin(\alpha )=2\sqrt{5}$$

The equation is in standard form.

$$25\sqrt{2}\sin(\alpha )y=2\sqrt{5}$$

Divide both sides by 25\sqrt{2}\sin(\alpha ).

$$\frac{25\sqrt{2}\sin(\alpha )y}{25\sqrt{2}\sin(\alpha )}=\frac{2\sqrt{5}}{25\sqrt{2}\sin(\alpha )}$$

Dividing by 25\sqrt{2}\sin(\alpha ) undoes the multiplication by 25\sqrt{2}\sin(\alpha ).

$$y=\frac{2\sqrt{5}}{25\sqrt{2}\sin(\alpha )}$$

Divide 2\sqrt{5} by 25\sqrt{2}\sin(\alpha ).

$$y=\frac{\sqrt{10}}{25\sin(\alpha )}$$

Solve for α

$$\alpha =arcSin(\frac{1}{25}y^{-1}\times 10^{\frac{1}{2}})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$\alpha =\pi +2\pi n_{2}+\left(-1\right)arcSin(\frac{1}{25}y^{-1}\times 10^{\frac{1}{2}})\text{, }n_{2}\in \mathrm{Z}$$