<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] Avoid squaring whenever possible as it immediately introduces 
 extraneous  root(s).
$$\sin x-\sqrt3\cos x=1$$
Method$\#1:$
Use Prosthaphaeresis Formulas
Method$\#2:$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
Method$\#3:$
Use Double Angle formula,
$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2$
and $1-\sin x=\left(\cos\dfrac x2-\sin\dfrac x2\right)^2$
We immediately have $\cos\dfrac x2-\sin\dfrac x2$ as common factor 
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