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

Solution :
We have,
$$\eqalign{ & 4\cos x\left( {2 - 3\,{{\sin }^2}x} \right) + \left( {\cos 2x + 1} \right) = 0 \cr & \Rightarrow 4\cos x\left( {3\,{{\cos }^2}x - 1} \right) + 2\,{\cos ^2}x = 0 \cr & \Rightarrow 2\cos x\left( {6\,{{\cos }^2}x + 2} \right)\left( {2\cos x - 1} \right) = 0 \cr & \Rightarrow 2\cos x\left( {3\cos x + 2} \right)\left( {2\cos x - 1} \right) = 0 \cr & \Rightarrow {\text{either}}\,\,\cos x = 0{\text{ which gives }}x = \frac{\pi }{2} \cr} $$
$${\text{or }}\cos x = - \frac{2}{3}$$
Which gives no value of $$x$$ for which $$0 \leqslant x \leqslant \frac{\pi }{2}$$   or $$\cos x = \frac{1}{2},$$   which gives $$x = \frac{\pi }{3}$$
So, the required difference $$ = \frac{\pi }{2} - \frac{\pi }{3} = \frac{\pi }{6}$$