<s>[INST] 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? 
 [/INST] 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
$$\boxed{x = -\frac{1}{2}\arccos\frac{4\sqrt 3 - 3}{13} + k\pi,\ \ k \in \Bbb Z} $$
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