<s>[INST] Is there any other method to solve $\sin x -\sqrt 3 \cos x=1$? Consider 
$$
\sin x -\sqrt 3 \cos x=1
$$
for $0\leq x \leq 2\pi$.
I solved it by converting $\sin x -\sqrt 3 \cos x$ to $2\sin(x-\pi/3)$ as follows
\begin{align*}
\sin x -\sqrt 3 \cos x &= r\sin(x-\theta)\\
&= r\sin x\cos\theta -r \cos x\sin\theta\\
\end{align*}
we have
\begin{align*}
r\cos \theta &= 1\\
r\sin \theta &=\sqrt 3
\end{align*}
where  $r=2$ and $\theta=\frac{\pi}{3}$ are the solution.
So the equation becomes 
$$
2\sin (x-\pi/3) =1
$$
and the solution are $x=\pi/2$ and $x=7\pi/6$.
Is there any other method to solve it?
 [/INST] You can convert it directly:
$$\sin x -\sqrt 3 \cos x=\sin x -\tan(\frac{\pi}{3}) \cos x=\sin x -\frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})} \cos x\\
=\frac{\sin x\cos(\frac{\pi}{3}) -\sin(\frac{\pi}{3})\cos x}{\cos(\frac{\pi}{3})} =\frac{\sin (x-\frac{\pi}{3})}{\cos(\frac{\pi}{3})} $$
In general, when dealing with $\sin(x)\pm A \cos(x)$, to convert it just write $A=\tan(\theta)$.
</s>