If x + frac{1}{x} = 2cos theta , then x is eq | vedclass

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(C) Given the equation $x + \frac{1}{x} = 2\cos 	heta$. Multiply both sides by $x$ to get $x^2 + 1 = 2x\cos 	heta$,which rearranges to $x^2 - 2x\cos

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$\cos 	heta \pm i\sin 	heta $

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(C) Given the equation $x + \frac{1}{x} = 2\cos 	heta$. Multiply both sides by $x$ to get $x^2 + 1 = 2x\cos 	heta$,which rearranges to $x^2 - 2x\cos 	heta + 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-2\cos 	heta, c=1$: $x = \frac{2\cos 	heta \pm \sqrt{4\cos^2 	heta - 4}}{2}$. $x = \frac{2\cos 	heta \pm \sqrt{4(\cos^2 	heta - 1)}}{2}$. Since $\cos^2 	heta - 1 = -\sin^2 	heta$,we have $x = \frac{2\cos 	heta \pm \sqrt{-4\sin^2 	heta}}{2}$. $x = \frac{2\cos 	heta \pm 2i\sin 	heta}{2} = \cos 	heta \pm i\sin 	heta$.

If $x + \frac{1}{x} = 2\cos \theta ,$ then $x$ is equal to

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