b) $2{\sin}^3x-\cos 2x+\cos x=0$
$\Rightarrow 2{\sin}^3x-(1-2{\sin}^2x)+\cos x=0$
$\Rightarrow 2{\sin}^3x-1+2{\sin}^2x+\cos x=0$
$\Rightarrow 2{\sin}^2x(\sin x+1)+(\cos x-1)=0$
$\Rightarrow2(1-{\cos}^2x)(\sin x+1)+(\cos x-1)=0$
$\Rightarrow 2(1+\cos x)(1-\cos x)(\sin x+1)+(\cos x-1)=0$
$\Rightarrow(\cos x-1)[1-2(1+\cos x)(\sin x+1)]=0$
$\Rightarrow (\cos x-1)(1-2\sin x-2-2\cos x\sin x-2\cos x)=0$
$\Rightarrow (\cos x-1)[-(\sin x+\cos x)^2-2(\sin x+\cos x)]=0$
$\Rightarrow -(\cos x-1)(\sin x+\cos x)(\sin x+\cos x+2)=0$
$\Rightarrow \left[ \begin{array}{l} \cos x-1=0 \\ \sin x+\cos x=0 \\\sin x+\cos x=-2\end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} \cos x=1 \\\sqrt2\sin( x+\dfrac{\pi}{4})=0 \\\sqrt2\sin( x+\dfrac{\pi}{4})=-2\end{array} \right .$
$\Rightarrow \left[ \begin{array}{l}x=k2\pi \\ \sin (x+\dfrac{\pi}{4})=0 \\\sin( x+\dfrac{\pi}{4})=-\dfrac{2}{\sqrt2}<-1(l)\end{array} \right .$
$\Rightarrow \left[ \begin{array}{l}x=k2\pi \\ x+\dfrac{\pi}{4}=k\pi \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l}x=k2\pi \\ x=-\dfrac{\pi}{4}+k\pi \end{array} \right .(k\in\mathbb Z)$
