Solve the cubic with roots expressed as tangents.

01 / Standard#Your turn

x^3-3\sqrt3x^2-3x+\sqrt3=0

Official paper · jm02-2023 · 4(b)(ii) · PDF 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Set x=tanθ with −π/2<θ<π/2.

Hint 2

Match the triple-angle tangent ratio to √3.

Worked solution

The potential denominator zeros x=±1/√3 are not roots of the cubic, so division is valid.
$\frac{3x-x^3}{1-3x^2}=\sqrt3\iff\tan3\theta=\sqrt3$
Solve and restrict θ to its selected interval.
$\theta=\pi/9+n\pi/3\implies\theta=-2\pi/9,\pi/9,4\pi/9$
Transform back to all three real roots.
$x=\tan(-2\pi/9),\ \tan(\pi/9),\ \tan(4\pi/9)$

tan(−2π/9), tan(π/9), tan(4π/9).

Checks and common pitfalls: Taking only the principal arctangent of √3 would miss two cubic roots.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The potential denominator zeros x=±1/√3 are not roots of the cubic, so division is valid.
$\frac{3x-x^3}{1-3x^2}=\sqrt3\iff\tan3\theta=\sqrt3$
Solve and restrict θ to its selected interval.
$\theta=\pi/9+n\pi/3\implies\theta=-2\pi/9,\pi/9,4\pi/9$
Transform back to all three real roots.
$x=\tan(-2\pi/9),\ \tan(\pi/9),\ \tan(4\pi/9)$

Think first. Reveal a hint when the class is ready.

2023 JM02