Solve equation(8cos3x+1)3=162cos⁡x−27

Question

Kathryn Duggan · Accepted Answer

This approach is based off knowing that the answers are

\frac{2 π}{9}, \frac{4 π}{9}, \frac{8 π}{9}

. Another equation whose solutions are exactly those values, is

2 \cos 3 x + 1 = 0

, so we'd look to factorize that out.
We know that

2 \cos 3 x + 1 = 8 \cos^{3} x - 6 \cos x + 1

, so

(8 \cos^{3} x + 1) = (2 \cos 3 x + 1 + 6 \cos x)

. Henceforth, let

w = 2 \cos 3 x + 1

We have

{(w + 6 \cos x)}^{3} = 162 \cos x - 27

, or that

w^{3} + 3 w^{26} \cos x + 3 w {(6 \cos x)}^{2} + 216 \cos^{3} x = 162 \cos x - 27 = - 27 w

hence

w^{3} + 18 w^{2} \cos x + 108 w \cos^{2} x + 27 w = 0

This factorizes into

0 = w (w^{2} + 18 w \cos x + 108 \cos^{2} x + 27)

= w [{(w + 9 \cos x)}^{2} + 27 \cos^{2} x + 27]

It is clear that the other term is strictly positive, so has no solution for x (regardless of what w is), so we must have

w = 0

. Hence, the solutions are

x = \frac{2 π}{9}, \frac{4 π}{9}, \frac{8 π}{9}

Solve equation (8\cos^3 x+1)^3=162\cos x-27