The sum of the roots of the equation  ${\sin }^2\theta -\cos \theta =\frac{1}{4} in [0,2\mathrm{\pi }] is$

${\sin }^2\theta -\cos \theta =\frac{1}{4}$

$\Rightarrow 1-{\cos }^2\theta -\cos \theta =\frac{1}{4}\Rightarrow {\cos }^2\theta +\cos \theta -\frac{3}{4}=0$

Solving, we get $\cos \theta =\frac{1}{2},-\frac{3}{2}$

$\Rightarrow \cos \theta \mid =\frac{1}{2}\left(\because \cos \theta \ne -\frac{3}{2}\right)\Rightarrow \theta =\frac{\pi }{3},\frac{5\pi }{3}$

$\therefore$Sum of the roots $=\frac{\pi }{3}+\frac{5\pi }{3}=2\pi$