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Try substitution: \( x = 2y \):
Let \( y = \cos\theta \), then:
4\cos^3\theta - 4\cos\theta + 1 = 0 \implies \cos 3\theta = \frac{1}{4}
Using identity: \( \cos 3\theta = \frac{1}{4} \implies 3\theta = \pm \arccos\left(\frac{1}{4}\right) + 2k\pi \)
\theta = \frac{1}{3} \arccos\left(\frac{1}{4}\right),\quad \frac{1}{3}\left(2\pi - \arccos\left(\frac{1}{4}\right)\right),\quad \frac{1}{3}\left(4\pi + \arccos\left(\frac{1}{4}\right)\right)
So the three real solutions for \( y \) are:
y_k = \frac{1}{3} \arccos\left(\frac{1}{4}\right) + \frac{2k\pi}{3},\quad k = 0,1,2
x_k = 2\cos\left( \frac{1}{3} \arccos\left(\frac{1}{4}\right) + \frac{2k\pi}{3} \right),\quad k = 0,1,2
But for simplicity and clarity in competition setting, and since no cleaner algebraic form exists, we conclude:
The real solutions are the three roots of \( x