If $1+{\sin }^2\theta =3\sin \theta .\cos \theta$ , then the solution set in $\left[0,\frac{\pi }{2}\right]$   is

Divide both sides with ${\cos }^2\theta$

${\sec }^2\theta +{\tan }^2\theta =3\tan \theta$

$2{\tan }^2\theta -3\tan \theta +1=0$

$\left(2\tan \theta -1\right)\left(\tan \theta -1\right)=0$

$\tan \theta =\frac{1}{2},\tan \theta =1$

$\therefore \left\{\frac{\pi }{4},{\tan }^{-1}\frac{1}{2}\right\}$