If 2 cos theta = x + frac{1}{x},then 2 cos 3 | vedclass

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(D) Given $2 \cos 	heta = x + \frac{1}{x}$,which implies $\cos 	heta = \frac{1}{2} \left(x + \frac{1}{x}ight)$.Using the triple angle identity $\c

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$x^3 + \frac{1}{x^3}$

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(D) Given $2 \cos 	heta = x + \frac{1}{x}$,which implies $\cos 	heta = \frac{1}{2} \left(x + \frac{1}{x}ight)$.Using the triple angle identity $\cos 3 	heta = 4 \cos^3 	heta - 3 \cos 	heta$,we have:$2 \cos 3 	heta = 2 [4 \cos^3 	heta - 3 \cos 	heta] = 8 \cos^3 	heta - 6 \cos 	heta$.Substitute $\cos 	heta = \frac{1}{2} \left(x + \frac{1}{x}ight)$:$2 \cos 3 	heta = 8 \left[ \frac{1}{2} \left(x + \frac{1}{x}ight) ight]^3 - 6 \left[ \frac{1}{2} \left(x + \frac{1}{x}ight) ight]$$= 8 \left[ \frac{1}{8} \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) ight) ight] - 3 \left(x + \frac{1}{x}ight)$$= \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) ight) - 3 \left(x + \frac{1}{x}ight)$$= x^3 + \frac{1}{x^3}$.

If $2 \cos \theta = x + \frac{1}{x}$,then $2 \cos 3 \theta = $

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