You may use the half angle identity to write `cos^2 2x ` such that:

`cos^2 2x = (1 + cos 4x)/2`

You may use the double angle identity to write `cos 8x` such that:

`cos 8x = cos 2(4x) = 2cos^2 4x - 1`

Hence, substituting `2cos^2 4x - 1` for `cos 8x ` and `(1 + cos 4x)/2` for `cos^2 2x` yields:

`2cos^2 4x - 1 = (1 + cos 4x)/2 `

`4cos^2 4x -2- 1- cos 4x = 0`

You should come up with the substitution such that:

`cos 4x = y `

`4y^2 - y - 3 = 0`

Using the quadratic formula yields:

`y_(1,2) = (1+-sqrt(1 + 48))/8`

`y_(1,2) = (1+-sqrt49)/8`

`y_(1,2) = (1+-7)/8 =gt y_1 = 1 ; y_2 = -6/8 =gt y_2 = -3/4`

You need to solve the equations `cos 4x = y_1` and `cos 4x = y_2` such that:

`cos 4x = 1 =gt 4x = 2npi =gt x = (npi)/2`

`cos 4x = -3/4 =gt 4x = +-cos^(-1)(-3/4) + 2npi`

`x = pi/4+-(cos^(-1)(3/4) )/4 + (npi)/2`

**Hence, evaluating the solutions to the given equation yields `x = (npi)/2` and `x = pi/4+-(cos^(-1)(3/4) )/4 + (npi)/2.` **