The least difference between the roots, in the first quadrant $\left(0\le x\le \frac{\pi }{2}\right)$ of the equation $4\cos x\left(2-3{\sin }^{2}x\right)+\left(\cos 2x+1\right)=0$, is

$We have 4\cos x\left(3{\cos }^{2}x-1\right)+2{\cos }^{2}x=0$

$\Rightarrow 2\cos x\left(3\cos x+2\right)\left(2\cos x-1\right)=0$

$$\begin{gathered} \Rightarrow \cos x=0 or \cos x=\frac{1}{2} \left(\because \cos x\ne -\frac{2}{3} as 0\le x\le \frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow x=\frac{\mathrm{\pi }}{2} or \frac{\mathrm{\pi }}{3} \\ \therefore the difference of values is \frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{6} \end{gathered}$$