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.

I continue from where you left off.
$$ 3\cos^2 x - 3\sin^2 x + 1 = 4\cos^2 x -\cos^2 x - \sin^2 x - 2\sin^2 x + 1 $$
$$ = 4\cos^2 x - 2\sin^2 x \boxed{-\cos^2 x - \sin^2 x + 1 } $$
$$ = 4\cos^2 x - 2\sin^2 x. $$
