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 $-2.414213562$
$$\cos x = 0.4142135624$$
$$x=1.4371774 , 5.139467567$$

Corrected solution:
Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{1-\cos^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \csc x (\cos^2 x+2\cos x-1)=0$$
$$\cot x =0 \rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$$
$$\csc x = 0 \rightarrow DNE$$
$$\cos^2 x +2\cos x -1=0 \rightarrow \cos x = \frac{-2\pm\sqrt{4-4*-1}}{2}=-1\pm\sqrt{2}$$
$$x=2\pi n \pm \cos^{-1} \left( -1\pm\sqrt{2}\right)$$
$$n \in \mathbb{Z}$$
