<s>[INST] How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\cos x(\cos x-\sqrt{3}\sin x)=0$$
$2\cos x=0\Rightarrow x\in \{\frac{\pi }2(2n-1):n\in\Bbb Z\}$
But how do I solve $$\cos x-\sqrt{3}\sin x=0$$
 [/INST] Hint :
$$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$
Note : You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in the final solution, as the $\cos$ zeros are no part of it.
</s>