Find $x$ if $\cot(x)=\csc(12^{\circ})-\sqrt{3}$ Find $x$ in degrees if $\cot(x)=\csc(12^{\circ})-\sqrt{3}$
My attempt:
$$\cot (x)=\frac{1}{\sin (12^{\circ})}-2 \sin \left(60^{\circ}\right)$$
$$\Rightarrow \cot x=\frac{1-2 \sin (12^{\circ}) \sin (60^{\circ})}{\sin \left(12^{\circ}\right)}$$
$$\Rightarrow \cot x=\frac{1-\cos 48^{\circ}+\cos 72^{\circ}}{\sin \left(12^{\circ}\right)}$$
Now let $, \theta=12^{\circ},s=\sin(\theta)$, then we get
$$\cot x=\frac{1-\cos (4 \theta)+\cos (6\theta)}{\sin (\theta)}$$
Converting to rational function in $s$, we get
$$\cot x=\frac{-32 s^{6}+40 s^{4}-10 s^{2}+1}{s}$$
 A: This is a very similar yet a little different question compared to this problem, and the answer by @albert chan paves the way for a similar solution to this problem.

$$ \sin(12°) = \sin(30°-18°) = \sin(30°)\cos(18°) - \cos(30°)\sin(18°)$$
$$= {1\over2} (\cos(18°) - \sqrt3 \sin(18°))$$
Converting $\sin$ to $\csc$ and rationalizing the denominator:
$$\csc(12°) = \left({2 \over \cos(18°) - \sqrt3 \sin(18°)}\right) \left({\cos(18°) + \sqrt3 \sin(18°) \over \cos(18°) + \sqrt3 \sin(18°)}\right)$$
$$= {2(\cos(18°) + \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} $$
Using $\cot(x)\sin(x)=\cos(x)$ for numerator and  Pythagorean trig identity for denominator:
$$= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) + \sqrt3) 
$$
Let $s=\sin(18°)$, using multiple angles formula and the fact that $s≠1$:
$$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s -1 = (s-1)(4s^2+2s-1)^2 = 0 $$
$$\implies 4s^2+2s-1 = 0 \implies {2s \over 1-4s^2} = 1$$
$$\fbox{$ \cot(x) = (\cot(18°) + \sqrt3) - \sqrt3 = \cot(18°)$}$$
