If sqrt{x} + frac{1}{sqrt{x}} = 2cos theta,th | vedclass

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(B) Given,$\sqrt{x} + \frac{1}{\sqrt{x}} = 2\cos 	heta$  $(i)$On squaring both sides,we get $x + \frac{1}{x} + 2 = 4\cos^2 	heta$$x + \frac{1}{x} =

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$2\cos 12	heta$

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(B) Given,$\sqrt{x} + \frac{1}{\sqrt{x}} = 2\cos 	heta$  $(i)$On squaring both sides,we get $x + \frac{1}{x} + 2 = 4\cos^2 	heta$$x + \frac{1}{x} = 4\cos^2 	heta - 2 = 2(2\cos^2 	heta - 1) = 2\cos 2	heta$  $(ii)$Again squaring both sides,$x^2 + \frac{1}{x^2} + 2 = 4\cos^2 2	heta$$x^2 + \frac{1}{x^2} = 4\cos^2 2	heta - 2 = 2(2\cos^2 2	heta - 1) = 2\cos 4	heta$  $(iii)$Now,taking the cube of both sides of $(iii)$:$(x^2 + \frac{1}{x^2})^3 = (2\cos 4	heta)^3$$x^6 + \frac{1}{x^6} + 3(x^2 \cdot \frac{1}{x^2})(x^2 + \frac{1}{x^2}) = 8\cos^3 4	heta$$x^6 + \frac{1}{x^6} + 3(2\cos 4	heta) = 8\cos^3 4	heta$$x^6 + \frac{1}{x^6} = 8\cos^3 4	heta - 6\cos 4	heta = 2(4\cos^3 4	heta - 3\cos 4	heta)$Using the identity $\cos 3A = 4\cos^3 A - 3\cos A$,we get:$x^6 + \frac{1}{x^6} = 2\cos 3(4	heta) = 2\cos 12	heta$.

If $\sqrt{x} + \frac{1}{\sqrt{x}} = 2\cos \theta$,then ${x^6} + {x^{-6}} = $

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