Problem Statement
Let $z = \cos\theta + i\sin\theta$.
(a) Use De Moivre's theorem to express $\cos(3\theta)$ in terms of $\cos\theta$. [4 marks]
(b) Hence, solve the equation $8x^3 - 6x - 1 = 0$ for $x \in \mathbb{R}$. [3 marks]
Verified Solution & Marking Scheme
Part (a): Expand (cos θ + i sin θ)³
$$\cos(3\theta) + i\sin(3\theta) = (\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$$
Equate real parts
$$\cos(3\theta) = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta = 4\cos^3\theta - 3\cos\theta$$ (AG)
Part (b): Substitute x = cos θ
$$8x^3 - 6x - 1 = 0 \implies 2(4x^3 - 3x) = 1 \implies 4x^3 - 3x = \frac{1}{2}$$. Substituting $$x = \cos\theta$$ yields $$\cos(3\theta) = \frac{1}{2}$$.
Solve for roots
$$3\theta = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3} \implies \theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}$$. Hence the three real roots are $$x = \cos\left(\frac{\pi}{9}\right), \cos\left(\frac{5\pi}{9}\right), \cos\left(\frac{7\pi}{9}\right)$$