Question: "Given cos⁡3θ=4(cos⁡θ)3−3cos⁡θ, solve 4x3−9x−1=0, correct to 3 decimal..."

sentidosin0q

Answered question

2022-01-26

Given

\cos 3 θ = 4 {(\cos θ)}^{3} - 3 \cos θ

, solve

4 x^{3} - 9 x - 1 = 0

, correct to 3 decimal places.

euromillionsna

Beginner2022-01-27Added 16 answers

Render

x = a \cos θ

. Thereby

4 x^{3} - 9 x = 4 a^{3} \cos^{3} θ - 9 a \cos θ

= a^{3} (4 \cos^{3} θ - (\frac{9}{a^{2}}) \cos θ)

If we put

a = \sqrt{3}

then

\frac{9}{a^{2}} = 3

and then

4 x^{3} - 9 x = 3 \sqrt{3} \cos (3 θ) = 1

where the "=1" cones from the constant term of the original equation. So

\cos (3 θ) = \frac{\sqrt{3}}{9}

, from which we can get values of

θ

to four decimal places. That should be enough accuracy as

x = a \cos θ = \sqrt{3} \cos θ

has a derivative within

\pm 2

for all

θ

Aiden Cooper

Beginner2022-01-28Added 14 answers

Let

x = a \cos θ

. We want

4 x^{3} - 9 x

as a multiple of

\cos 3 θ

; that is to say

\frac{4 a^{3}}{9 a} = \frac{43}{}

, so

a = \sqrt{3}

.
Thus, we find

3 \sqrt{3} \cos 3 θ = 1

and then

x = \sqrt{3} \cos (\frac{13}{\arccos \frac{1}{3 \sqrt{3}}})

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