Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$ I'm stuck on a trigonometry question. The formulae I have been given don't seem to help so I had to come here.
The question is:

Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$

If these formulae can be used I would be glad if you could show me how:


*

*$\cos(2x) = 2\cos^2(x) - 1$ 

*$\cos(2x) = 1 - 2\sin^2(x)$

*$\cos^2(x) = \frac 12(1 + \cos(2x))$

*$\sin^2(x) = \frac 12(1 - \cos(2x))$


In the right side you can use:
\begin{equation}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\\
\end{equation}
In this case:
\begin{equation}
\cos(x-\frac{\pi}{6})=\cos x\cos\frac{\pi}{6}+\sin x\sin\frac{\pi}{6}\\
\cos(x-\frac{\pi}{6})=\frac{\sqrt 3}{2}\cos x+\frac{1}{2}\sin x
\end{equation}
