If x + frac{1}{x} = sqrt{3}, then x = | vedclass

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(D) Given $x + \frac{1}{x} = \sqrt{3}$.Multiplying by $x$,we get $x^2 - \sqrt{3}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4

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$\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$

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(D) Given $x + \frac{1}{x} = \sqrt{3}$.Multiplying by $x$,we get $x^2 - \sqrt{3}x + 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:$x = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4(1)(1)}}{2(1)}$$x = \frac{\sqrt{3} \pm \sqrt{3 - 4}}{2} = \frac{\sqrt{3} \pm \sqrt{-1}}{2} = \frac{\sqrt{3} \pm i}{2}$.Thus,$x = \frac{\sqrt{3}}{2} \pm \frac{i}{2}$.Taking the positive sign,$x = \frac{\sqrt{3}}{2} + i \frac{1}{2} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$.

If $x + \frac{1}{x} = \sqrt{3},$ then $x =$

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