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.

You know that
\begin{equation}
\textrm{sin}^2x + \textrm{sin}^2 2x + \textrm{sin}^2 3x = 1 \qquad (\textrm{Eq.} 1)
\end{equation}
Notice that 
\begin{equation}
\begin{aligned}
\textrm{sin}^2x + \textrm{cos}^2 x = 1 \Rightarrow \textrm{sin}^2x = 1 - \textrm{cos}^2 x  
\end{aligned}
\end{equation}
Substituting the identity in Equation (1):
\begin{equation}
\begin{aligned}
\underbrace{\textrm{sin}^2x}_{1-\textrm{cos}^2x} &  + \underbrace{\textrm{sin}^2 2x}_{1-\textrm{cos}^2 2x } + \underbrace{\textrm{sin}^2 3x}_{1-\textrm{cos}^2 3x} = 1 \\ \\ & \Rightarrow 1-\textrm{cos}^2x + 1-\textrm{cos}^2 2x + 1-\textrm{cos}^2 3x = 1 \\ & \Rightarrow 3 -\left( \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x \right) = 1 \\ & \Rightarrow 2 = \textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x
\end{aligned}
\end{equation} 
Therefore: 
\begin{equation}
\textrm{cos}^2x + \textrm{cos}^2 2x + \textrm{cos}^2 3x = 2
\end{equation}
