Solve sqrt{3}sinalpha=2kcosalpha+2

$$\sqrt { 3 } \sin \alpha = 2 k \cos \alpha + 2$$

Solve for k

$$k=-\frac{-\sqrt{3}\sin(\alpha )+2}{2\cos(\alpha )}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}+\frac{\pi }{2}$$

Steps for Solving Linear Equation

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

$$2k\cos(\alpha )+2=\sqrt{3}\sin(\alpha )$$

Subtract 2 from both sides.

$$2k\cos(\alpha )=\sqrt{3}\sin(\alpha )-2$$

The equation is in standard form.

$$2\cos(\alpha )k=\sqrt{3}\sin(\alpha )-2$$

Divide both sides by 2\cos(\alpha ).

$$\frac{2\cos(\alpha )k}{2\cos(\alpha )}=\frac{\sqrt{3}\sin(\alpha )-2}{2\cos(\alpha )}$$

Dividing by 2\cos(\alpha ) undoes the multiplication by 2\cos(\alpha ).

$$k=\frac{\sqrt{3}\sin(\alpha )-2}{2\cos(\alpha )}$$