4cos2x – 8cosx – 1 = 0

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To solve this trigonometric equation, we can use the double angle formula for cosine which states that cos(2x) = 2cos^2(x) - 1.

Substitute this formula into the equation:

4(2cos^2(x) - 1) - 8cos(x) - 1 = 0
8cos^2(x) - 4 - 8cos(x) - 1 = 0
8cos^2(x) - 8cos(x) - 5 = 0

Now, let y = cos(x). The equation becomes:

8y^2 - 8y - 5 = 0

Now we have a quadratic equation that we can solve using the quadratic formula:

y = [8 ± √(8^2 - 48(-5))] / 2*8
y = [8 ± √(64 + 160)] / 16
y = [8 ± √224] / 16

y = [8 ± 14.97] / 16

Now we solve for y:

y1 = (8 + 14.97) / 16 ≈ 1.187
y2 = (8 - 14.97) / 16 ≈ -0.874

Since y = cos(x), we can find the corresponding values of x:

x1 = cos^(-1)(1.187) is not valid
x2 = cos^(-1)(-0.874) ≈ 2.610

Therefore, the solutions to the equation 4cos(2x) - 8cos(x) - 1 = 0 are x ≈ 2.610 + 2πn, n ∈ ℤ.