Question

[Sqrt(32cos^2x+1)-4]/[4-sqrt(32sin^2x+1)]=1

Answer 1

what happened to last question??

Answer 2

It was a mess. Made it more clear.

Answer 3

well ... try it. put sin^2 x = u and cos^2 u = 1-x

Answer 4

Alright, just a sec.

Answer 5

cos^2u =1-u, not 1-x?

Answer 6

sorry cos^2 x = 1 - u

Answer 7

Something went wrong, trying again.

Answer 8

\[ \frac{\sqrt{32(1-u)+1}}{4-\sqrt{32u+1}} - 1 = 0\]

Answer 9

\[\frac{\sqrt{32cos^2x+1}-4}{4-\sqrt{32sin^2x+1}}=1\]\[\sqrt{32cos^2x+1}-4=4-\sqrt{32sin^2x+1}\]\[\sqrt{32cos^2x+1} + \sqrt{32sin^2x+1}=8\]\[(\sqrt{32cos^2x+1} + \sqrt{32(1-cos^2x)+1})^2=64\]\[(\sqrt{32cos^2x+1})^2+ 2(\sqrt{32cos^2x+1})(\sqrt{33-32cos^2x})\]\[+(\sqrt{33-32cos^2x})^2=64\]\[(\sqrt{32cos^2x+1} + \sqrt{32(1-cos^2x+1})^2=64\]\[32cos^2x+1+ 2(\sqrt{32cos^2x+1})(\sqrt{33-32cos^2x})+33-32cos^2x=64\]\[ 2(\sqrt{(32cos^2x+1)(33-32cos^2x)})+34=64\]\[ 2(\sqrt{(32cos^2x+1)(33-32cos^2x)})=30\]\[ \sqrt{(32cos^2x+1)(33-32cos^2x)}=15\]\[ \sqrt{(32cos^2x+1)(33-32cos^2x)}^2=15^2\]\[ (32cos^2x+1)(33-32cos^2x)=225\]
Then, it's a quadratic...

Answer 10

Not sure if it's right though. May have made mistakes there, but the concept is like that :|

Answer 11

Thanks!

Answer 12

welcome, hope it helps :)