Solve 6cos^2x=4-si

$$6 \cos ^ { 2 } x = 4 - s i$$

Solve for s

$$s=2\left(3i\left(\cos(x)\right)^{2}-2i\right)$$

Steps for Solving Linear Equation

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

$$4-si=6\left(\cos(x)\right)^{2}$$

Multiply -1 and i to get -i.

$$4-is=6\left(\cos(x)\right)^{2}$$

Subtract 4 from both sides.

$$-is=6\left(\cos(x)\right)^{2}-4$$

Divide both sides by -i.

$$\frac{-is}{-i}=\frac{2\left(3\left(\cos(x)\right)^{2}-2\right)}{-i}$$

Dividing by -i undoes the multiplication by -i.

$$s=\frac{2\left(3\left(\cos(x)\right)^{2}-2\right)}{-i}$$

Divide 2\left(3\left(\cos(x)\right)^{2}-2\right) by -i.

$$s=2\left(3i\left(\cos(x)\right)^{2}-2i\right)$$

Solve for x

$$x=\frac{1}{2}i\ln(6)+\left(-i\right)\ln(\left(4+\left(-i\right)s\right)^{\frac{1}{2}}+\left(-2+\left(-i\right)s\right)^{\frac{1}{2}})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}$$

$$x=\left(-\frac{1}{2}i\right)\ln(6)+\left(-i\right)\ln(\frac{1}{6}\left(4+\left(-i\right)s\right)^{\frac{1}{2}}+\left(-\frac{1}{6}\right)\left(-2+\left(-i\right)s\right)^{\frac{1}{2}})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}$$

$$x=\frac{1}{2}i\ln(6)+\left(-i\right)\ln(\left(-1\right)\left(4+\left(-i\right)s\right)^{\frac{1}{2}}+\left(-2+\left(-i\right)s\right)^{\frac{1}{2}})+2\pi n_{14}\text{, }n_{14}\in \mathrm{Z}$$

$$x=\left(-\frac{1}{2}i\right)\ln(6)+\left(-i\right)\ln(\left(-\frac{1}{6}\right)\left(4+\left(-i\right)s\right)^{\frac{1}{2}}+\left(-\frac{1}{6}\right)\left(-2+\left(-i\right)s\right)^{\frac{1}{2}})+2\pi n_{13}\text{, }n_{13}\in \mathrm{Z}$$