Question

Using the rule that cos3θ = 4(cosθ)^3 − 3 cosθ, show that cos 2π/9 is a root of the equation 8x^3 − 6x + 1 = 0

Answer 1

Below in bold.

Explanation:

Let x = cosθ, then

8(cosθ)^3 − 6cosθ + 1 = 0

---> 2(4(cosθ)^3 − 3 cosθ) + 1 = 0

---> 2(cos3θ) + 1 = 0

---> cos3θ = -1/2

---> θ = 2π/9

Therefore cos θ = = cos(2π/9) = x, and

cos(2π/9) is a root of the given eqation.