If \(x^2 - 2x\cos\theta + 1 = 0\), then what is the magnitude of \(x\)?
\(1\)
Solving \(x^2-2x\cos\theta+1=0\) by the quadratic formula gives \(x=\cos\theta \pm i\sin\theta\), since \(\cos^2\theta-1=-\sin^2\theta\). The magnitude is \(|x|=\sqrt{\cos^2\theta+\sin^2\theta}=1\).
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