If frac{1}{x} + x = 2cos theta, then {x^n} + | vedclass

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(A) Given $\frac{1}{x} + x = 2\cos 	heta$. Multiplying by $x$,we get $x^2 - 2x\cos 	heta + 1 = 0$. Using the quadratic formula,$x = \frac{2\cos 	he

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$2\cos n	heta$

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(A) Given $\frac{1}{x} + x = 2\cos 	heta$. Multiplying by $x$,we get $x^2 - 2x\cos 	heta + 1 = 0$. Using the quadratic formula,$x = \frac{2\cos 	heta \pm \sqrt{4\cos^2 	heta - 4}}{2} = \cos 	heta \pm i\sin 	heta$. By De Moivre's Theorem,$x^n = (\cos 	heta \pm i\sin 	heta)^n = \cos n	heta \pm i\sin n	heta$. Similarly,$\frac{1}{x^n} = x^{-n} = (\cos 	heta \pm i\sin 	heta)^{-n} = \cos n	heta \mp i\sin n	heta$. Adding these,we get $x^n + \frac{1}{x^n} = (\cos n	heta \pm i\sin n	heta) + (\cos n	heta \mp i\sin n	heta) = 2\cos n	heta$.

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

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