If \[\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta ,\]then \[{{x}^{6}}+{{x}^{-6}}=\] [Karnataka CET 2003]

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

1.	If \[\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta ,\]then \[{{x}^{6}}+{{x}^{-6}}=\] [Karnataka CET 2003]
A.	\[2\cos 6\theta \]
B.	\[2\cos 12\theta \]
C.	\[2\cos 3\theta \]
D.	\[2\sin 3\theta \]
Answer» C. \[2\cos 3\theta \]

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If \[\sqrt{x}+\frac{1}{\sqrt{x}}=2\cos \theta ,\]then \[{{x}^{6}}+{{x}^{-6}}=\] [Karnataka CET 2003]

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