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?

Squaring you have:
$$\sin^2 x + 3\cos^2x - 2\sqrt{3}\sin x\cos x = 1$$
So, $\cos^2 x = \sqrt{3}\sin x\cos x$
i.e., either $\cos x = 0$, in which case $x = \pi/2$ (since $3\pi/2$ is not a solution) or $\tan x= 1/\sqrt{3}$, in which case $x = 7\pi/6$
