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Home » So \( racz + wz - w = 1 \Rightarrow z + w = z - w \Rightarrow - Abu Waleed Tea

So \( racz + wz - w = 1 \Rightarrow z + w = z - w \Rightarrow - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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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 \...
\theta = \frac{1}{3} \arccos\left(\frac{1}{4}\right),\quad \frac{1}{3}...
So the three real solutions for \( y \) are:
y_k = \frac{1}{3} \arccos\left(\frac{1}{4}\right) + \frac{2k\pi}{3},\q...
x_k = 2\cos\left( \frac{1}{3} \arccos\left(\frac{1}{4}\right) + \frac{...
But for simplicity and clarity in competition setting, and since no cl...
The real solutions are the three roots of \( x

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