JEE Math Practice Question

If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation $$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$, with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :

1. A.$$\frac{2}{\sqrt{3}} e^{-2 \pi / 3}$$
2. B.$$\frac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}$$
3. C.$$\frac{1}{\sqrt{3}} e^{-2 \pi / 3}$$
4. D.$$\frac{1}{\sqrt{3}} e^{2 \pi / 3}$$