Difficult equation

[Math Tutor]

[tex]x(3-x)^2=1[/tex]
Find the roots of the equation.
Express its roots by trigonometric functions.
Author: r2d2.

[Guest]

Spoiler: show

Substitute [tex]x = 2\cos\theta +2[/tex] to get
[tex]x(3-x)^2 = (2\cos\theta +2)(1-2\cos\theta)^2 = 2-6\cos\theta+8\cos^3\theta = 2 + 2\cos 3\theta.[/tex]
So [tex]\cos 3\theta = -1/2[/tex] hence [tex]\theta = 40 + 120n[/tex] or [tex]80+120n[/tex] and therefore
[tex]x = 2\cos 40 +2[/tex], [tex]2\cos 80 + 2[/tex], [tex]2\cos 160 + 2[/tex].

R. Baber.

[Guest]

I could not think of a solution.