JEE Math Practice Question

Let y = y(x) be the solution of the differential equation $$\sin x{{dy} \over {dx}} + y\cos x = 4x$$, $$x \in \left( {0,\pi } \right)$$. If $$y\left( {{\pi \over 2}} \right) = 0$$, then $$y\left( {{\pi \over 6}} \right)$$ is equal to :

1. A.$$ - {4 \over 9}{\pi ^2}$$
2. B.$${4 \over {9\sqrt 3 }}{\pi ^2}$$
3. C.$$ - {8 \over {9\sqrt 3 }}{\pi ^2}$$
4. D.$$ - {8 \over 9}{\pi ^2}$$