How can I find $\cos(\theta)$ with $\sin(\theta)$? 
If $\sin^2x$ + $\sin^22x$ + $\sin^23x$ = 1, what does  $\cos^2x$ + $\cos^22x$ + $\cos^23x$ equal?

My attempted (and incorrect) solution:


*

*$\sin^2x$ + $\sin^22x$ + $\sin^23x$ = $\sin^26x$ = 1

*$\sin^2x = 1/6$

*$\sin x = 1/\sqrt{6}$

*$\sin x =$ opposite/hypotenuse

*Therefore, opp = 1, hyp = $\sqrt6$, adj = $\sqrt5$ (pythagoras theorum)

*$\cos x$ = adjacent/hypotenuse = $\sqrt5/\sqrt6$

*$\cos^2x$ = 5/6

*$\cos^2x + \cos^22x + \cos^23x = \cos^26x = 5/6 * 6 = 5$


I think I did something incorrectly right off the bat at step-2, and am hoping somebody will point me in the right direction.

$\sin^2x+\sin^22x+\sin^23x=1\implies(1-cos^2x)+ (1-\cos^22x) + (1-cos^23x)=1$
Rearranging terms we get $\cos^2x+\cos^22x+\cos^23x=2$
