Solve (4)y=2cos^2x/2-1

$$( 4 ) y = 2 \cos ^ { 2 } \frac { x } { 2 } - 1$$

Solve for y

$$y=\frac{\cos(x)}{4}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$4y=\cos(2\times \frac{x}{2})$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{\cos(x)}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{\cos(x)}{4}$$

Solve for x (complex solution)

$$x=i\ln(2)+\left(-2i\right)\ln(\left(4y+1\right)^{\frac{1}{2}}+\left(4y-1\right)^{\frac{1}{2}})+4\pi n_{5}\text{, }n_{5}\in \mathrm{Z}$$

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

$$x=i\ln(2)+\left(-2i\right)\ln(\left(-1\right)\left(4y+1\right)^{\frac{1}{2}}+\left(-1+4y\right)^{\frac{1}{2}})+4\pi n_{17}\text{, }n_{17}\in \mathrm{Z}$$

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

Solve for x

$$x=2\left(\arccos(\frac{\sqrt{2\left(4y+1\right)}}{2})+\pi n_{1}-\pi \right)\text{, }n_{1}\in \mathrm{Z}$$

$$x=2\left(-\arccos(\frac{\sqrt{2\left(4y+1\right)}}{2})+\pi n_{2}\right)\text{, }n_{2}\in \mathrm{Z}\text{, }|y|\leq \frac{1}{4}$$