# Use the result, cos3θ=4cos³θ-3cosθ, to solve completely the equation 8x³-6x+1?

Given answer is x=0.9397, -0.1737, -0.7660

### 2 Answers

- kbLv 76 months agoFavorite Answer
Assuming that you want to solve 8x³ - 6x + 1 = 0:

I believe you have the signs of the answers the other way around:

https://www.wolframalpha.com/input/?i=8x%C2%B3+-+6...

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Let x = cos θ.

Then, the equation transforms to 8 cos³θ - 6 cos θ + 1 = 0.

==> 2(4 cos³θ - 3 cos θ) + 1 = 0.

==> 2 cos(3θ) + 1 = 0, by using the given trig. identity.

Solve for θ:

cos(3θ) = -1/2

3θ = 2π/3 + 2πk or 4π/3 + 2πk, where k is any integer

==> θ = 2π/9 + 2πk/3 or 4π/9 + 2πk/3, where k = 0, 1, 2 (other roots are repeats of these values).

Finally, we solve for x using x = cos θ:

x = cos(2π/9) ≈ 0.7660

x = cos(2π/9 + 2π/3) ≈ -0.9397

x = cos(2π/9 + 4π/3) ≈ 0.1736

x = cos(4π/9) ≈ 0.1736

x = cos(4π/9 + 2π/3) ≈ -0.9397

x = cos(4π/9 + 4π/3) ≈ 0.7660.

Hence, the three distinct answers are x ≈ 0.7660, -0.9397, 0.1736.

(This makes sense, because the equation we solved has degree 3.)

I hope this helps!

- 6 months ago
cos(t) = x

8x^3 - 6x + 1 = 0

8cos(t)^3 - 6 * cos(t) + 1 = 0

2 * (4 * cos(t)^3 - 3 * cos(t)) + 1 = 0

2 * (4 * cos(t)^3 - 3 * cos(t)) = -1

4 * cos(t)^3 - 3 * cos(t) = -1/2

cos(3t) = -1/2

cos(3t) = cos(2pi/3 + 2pi * k) , cos(4pi/3 + 2pi * k)

3t = 2pi/3 + 2pi * k , 4pi/3 + 2pi * k

t = 2pi/9 + (2pi/3) * k , (4pi/9) + (2pi/3) * k

t = (2pi/9) + (6pi/9) * k , (4pi/9) + (6pi/9) * k

Assuming we want all values between t = 0 and t = 2pi

t = 2pi/9 , 8pi/9 , 14pi/9 , 4pi/9 , 10pi/9 , 16pi/9

t = 2pi/9 , 4pi/9 , 8pi/9 , 10pi/9 , 14pi/9 , 16pi/9

cos(t) = x

cos(2pi/9) , cos(4pi/9) , cos(8pi/9) , cos(10pi/9) , cos(14pi/9) , cos(16pi/9)

cos(16pi/9) = cos(2pi/9)

cos(14pi/9) = cos(4pi/9)

cos(10pi/9) = cos(8pi/9)

cos(2pi/9) , cos(4pi/9) , cos(8pi/9)

cos(2pi/9) = (1/4) * ((-4 + 4 * sqrt(3) * i)^(1/3) + (-4 - 4 * sqrt(3) * i)^(1/3))

cos(4pi/9) = 2 * cos(2pi/9)^2 - 1

cos(8pi/9) = 2 * cos(4pi/9)^2 - 1