Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$ Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$
Attempt.
I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solving it but its not proofing. I've tried replacing with trigonometric identities and simplifying. Please someone should help even if it is how to start?

The first thing to remember is
$$
\sin2x=2\sin x\cos x
$$
that settles the $-\sqrt{3}\sin2x$ part. Thus you just need to massage
$$
5\cos^2x+3\sin^2x
$$
recalling $\cos2x=\cos^2x-\sin^2x$; so rewrite it as
$$
\cos^2x+4\cos^2x-\sin^2x+4\sin^2x=\cos2x+4(\cos^2x+\sin^2x)
$$
You can also equivalently prove that
$$
5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x - \cos 2x + \sqrt{3}\sin2x - 4=0
$$
that goes like
\begin{multline}
5\cos^2x+3\sin^2x-\cos^2x+\sin^2x-4-\sqrt{3}(2\sin x\cos x-\sin2x)\\
=4\cos^2x+4\sin^2x-4-\sqrt{3}(\sin2x-\sin2x)=0
\end{multline}
