<s>[INST] Is this correct $(\cot x)(\sin x)=2(\cot x)^2$ $0≤x≤2\pi$ $x= \pi/2 , 3\pi/2$ $1.4371774 , 5.139467567$ Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{1-\cos^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \csc x (\cos x+2.414213562)(\cos -0.4142135624)=0$$
$$\cot x =0 \rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$$
$\csc x$ cannot be $0$
$\cos x$ cannot be $-2.414213562$
$$\cos x = 0.4142135624$$
$$x=1.4371774 , 5.139467567$$
 [/INST] i will use $$x = \cos t, y = \sin t , x^2 + y^2 = 1.$$  you have $$\cot t\sin t = 2 \cot^2 t \to \frac x y y=2\frac {x^2}{y^2} \to 0=x(2x-y^2)=x(-x^2 + 2x  + 1)$$ the solutions are $$x = 0, x = \frac{-2 \pm 2\sqrt 2}{-2} =1+\sqrt 2, \sqrt 2 - 1$$  since $x < 1,$ we have $$ x= 0, x = \sqrt 2 - 1$$
and the corresponding $t$ values are $$\pi/2, \cos^{-1}(\sqrt 2 - 1), 2\pi - \cos^{-1}(\sqrt 2 - 1), 3\pi/2.$$
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