$\frac{dy}{dx}-3\cot x.y=\sin 2x$
$I.F=e^{\underset{ }{\int }-3\cot xdx}=\frac{1}{{\sin }^{3}x}$

Solution is $y.\frac{1}{{\sin }^{3}x}=\int \sin 2x.\frac{1}{{\sin }^{3}x}dx$
$\frac{y}{{\sin }^{3}x}=-2\cos ecx+x---\left(1\right)$
But $y\left(\frac{\pi }{2}\right)=2\Rightarrow c=4$
$\therefore \frac{y}{{\sin }^{3}x}=-\cos ec+4$
$y\left(\frac{\pi }{6}\right)=0$