Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$ Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$
Using the formula $\cos 2x = \cos x - \sin^2 x$
I can say $3\cos 2x + 1 = 3(\cos^2 x - \sin^2 x) + 1$
$\Rightarrow 3\cos x^2 - 3\sin^2 x + 1$
But from there I don't see how I can get the answer.

You are very close. All you need to do is use the identity $\sin^2{x} + \cos^2{x} = 1$:
$\begin{eqnarray}3\cos{2x} + 1 & = & 3(\cos^2{x}-\sin^2{x}) + 1 \\
& = & 3(\cos^2{x}-\sin^2{x}) + \sin^2{x} + \cos^2{x} \\
& = & 4\cos^2{x} - 2\sin^2{x}
\end{eqnarray}$
QED.
Alternatively, if you don't feel comfortable turning the $1$ into $\sin^2{x}+\cos^2{x}$ (and trust me, it's a perfectly valid thing to do), you can equivalently break out one of the existing $\sin^2{x}$s and replace it with $1 - \cos^2{x}$, and as long as you get all the signs correct it amounts to the same thing.
