x = 2\cos\left(\frac{\pi}{9}\right),\quad 2\cos\left(\frac{5\pi}{9}\right),\quad 2\cos\left(\frac{7\pi}{9}\right)

February 22, 2026

$x = 2\cos\left(\frac{\pi}{9}\right),\quad 2\cos\left(\frac{5\pi}{9}\right),\quad 2\cos\left(\frac{7\pi}{9}\right)$

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After analyzing, the equation \( x^3 - 4x + 2 = 0 \) has **three real roots**, but none are rational. Using Cardano’s method or numerical approximation, we find:
Using precise computation:
The real solutions are:
But since the question asks for real solutions and the polynomial is cubic, we verify via direct root-finding:
Actually, applying the cubic formula or checking discriminant \( \Delta = -4(-4)^3 - 27(2)^2 = 256 - 108 = 148 > 0 \), which implies **three distinct real roots**.
Thus, the equation \( x^3 - 4x + 2 = 0 \) has three real solutions, expressible in trigonometric form or as approximations.

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