Question

Okay. A nice one.
Refer to post.

Answer 1

If \[\sqrt{x} + 1/\sqrt{x} = 2\cos(\theta) \]
Find \[x^6 + 1/x^6\]

Answer 2

\[\sqrt{x}=y\]
your equation is

Answer 3

\[y^2-2ycos \theta+1=0\]

Answer 4

solve for y find x

Answer 5

\[y=\frac{ -b \pm \sqrt{b^2-4*a*c} }{ 2*a }\] a=1, b= -2 cos theta and c=1

Answer 6

going with\[\sqrt{x}=e^{i\theta}\]\[x^6+x^{-6}=e^{12i\theta}+e^{-12i\theta}=2 \cos 12\theta\]

Answer 7

@mukushla Precisely the best way to do this. Well done.

Answer 8

thank u