Strategy for solving $\sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1$ I believe the way to solve the following equation is to use the "R-formula":
$$ \sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1 $$
If so, it should be rewritten as:
$ R\cos(x-\mathcal{L})$ or $R(\cos x\cos\mathcal{L}+\sin x\sin\mathcal{L})$.
With coefficients equated as:
$ \sqrt{3} = R\cos\mathcal{L} $
$ ? = R\sin x $
In the top equation, it looks like $ \cos x $ is the coefficient of $ \sin x $. But that doesn't work with R-formula (does it?).
I have spent some time trying to use double-angle identities for $\cos(2x)$ to remove something from the $\sin(x)\cos(x)$ expression, but am uncertain whether that's possible due to the leading $\sqrt{3}$.
I have also taken several suggestions from threads like this one, but repeatedly end up uncertain due to the same double-angle issue and/or coefficient issue.
Is R-formula the right approach to solving this equation? Should a double-angle identity be used first? Or is an entirely different approach correct? 

One more answer won't do any harm, hopefully. Here's another way to look at this "thing".
Once you replaced $\sin x \cos x$ with $\frac{1}{2}\sin 2x$, your equation becomes
$$\sqrt 3 \cos 2x - \frac{1}{2}\sin 2x = 1.$$
Recalling that cosine and sine are just abscissa and ordinate of points on the circumference of radius $1$, you can find the solutions by intersecting the line
$$r: \sqrt 3 X - \frac{1}{2}Y = 1$$
with the circumference
$$X^2 + Y^2 = 1.$$
This leads to the system of equations
$$
\begin{cases}
\sqrt 3 X - \frac{1}{2}Y = 1\\
X^2 + Y^2 = 1.
\end{cases}
$$

Replacing $Y = 2\sqrt 3 X - 2$ in the second equation yields the quadratic equation
$$13X^2-8\sqrt 3 X +3 =0$$
with solutions
$$ \cos 2x = X = \frac{4\sqrt 3 \pm 3}{13}.$$
Since solutions in terms of $2x$ are in first and fourth quadrant we can write the first set of solutions as
$$2x = \arccos\frac{4\sqrt 3 + 3}{13} + 2k\pi, $$
that is
$$\boxed{x =  \frac{1}{2}\arccos\frac{4\sqrt 3 + 3}{13} + k\pi, \ \ k \in \Bbb Z}$$
and the second set of solutions as
$$2x = -\arccos\frac{4\sqrt 3 - 3}{13} + 2k\pi,$$
that is
$$\boxed{x = -\frac{1}{2}\arccos\frac{4\sqrt 3 - 3}{13} + k\pi,\ \ k \in \Bbb Z} $$
