Determine the general solution of 8 cos^2x - 4cosx - 1 =0

(8cos^2x - 4cosx)/8 =1/8
cos^2x - 4cosx=1/8
cos^2x/cosx =1+4cosx/cosx
cosx = 5
x = cos^-1(5)
x = error

no, no. It's just a quadratic equation, in cosx, rather than just x

using the quadratic formula, we have

cosx = (1±√3)/4 or, -0.183, 0.683

So, just find the angles with those values as cosine.

To determine the general solution of the given equation 8cos^2x - 4cosx - 1 = 0, we can solve it using the quadratic formula.

Consider the equation in the form of a quadratic equation: 8cos^2x - 4cosx - 1 = 0

Using the quadratic formula, we have: x = (-B ± √(B^2 - 4AC)) / (2A)

Here, A = 8, B = -4, and C = -1.

Substituting the values, we get: x = (-(-4) ± √((-4)^2 - 4 * 8 * (-1))) / (2 * 8)

Simplifying further, we have: x = (4 ± √(16 + 32)) / 16
x = (4 ± √48) / 16
x = (4 ± 4√3) / 16
x = (1 ± √3)/4

Therefore, the general solution of the equation 8cos^2x - 4cosx - 1 = 0 is:

x = (1 + √3)/4, (1 - √3)/4