<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] We know that $\cos{\pi\over3}={1\over2}$ and $\sin{\pi\over3}={\sqrt3\over2}$, so
$$\begin{align}
\sin x-\sqrt3\cos x=1&\implies{1\over2}\sin x-{\sqrt3\over2}\cos x={1\over2}\\
&\implies\sin x\cos{\pi\over3}-\cos x\sin{\pi\over3}={1\over2}\\
&\implies\sin(x-{\pi\over3})={1\over2}\\
&\implies x-{\pi\over3}={\pi\over6}\quad\text{or}\quad x-{\pi\over3}=\pi-{\pi\over6}\\
&\implies x={\pi\over2}\quad\text{or}\quad x={7\pi\over6}
\end{align}$$
</s>