If x is real and x+frac{1}{x}=2 cos theta, th | vedclass

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(C) Given the equation $x+\frac{1}{x}=2 \cos 	heta$.Multiplying by $x$,we get $x^2 - 2x \cos 	heta + 1 = 0$.Since $x$ is a real number,the discrimin

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$\pm 1$

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(C) Given the equation $x+\frac{1}{x}=2 \cos 	heta$.Multiplying by $x$,we get $x^2 - 2x \cos 	heta + 1 = 0$.Since $x$ is a real number,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.$D = b^2 - 4ac = (-2 \cos 	heta)^2 - 4(1)(1) = 4 \cos^2 	heta - 4$.Setting $D \geq 0$,we have $4 \cos^2 	heta - 4 \geq 0$,which implies $\cos^2 	heta \geq 1$.Since the range of $\cos 	heta$ is $[-1, 1]$,the only possible values satisfying $\cos^2 	heta \geq 1$ are $\cos 	heta = 1$ or $\cos 	heta = -1$.Thus,$\cos 	heta = \pm 1$.

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

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