Given,

sin x $${{dy} \over {dx}} + y\cos y = 4x$$

$$ \Rightarrow \,\,\,\,{{dy} \over {dx}}\,$$ + y cot x = 4x cosec x

This is a linear differential equation of form,

$${{dy} \over {dx}}\,$$ + py = Q

Where p = cot x and Q = 4x cosec x

So, Integrating factor (I. F)

= $${e^{\int {pdx} }}$$

= $${e^{\int {\cot dx} }}$$

= $${e^{\ln \left| {\sin \,x} \right|}}$$

= sin x as $$x \in \left( {0,\pi } \right)$$

Solution of the differential equation is

y sin x = $$\int {} $$ 4x cosecx sinx dx + c

$$ \Rightarrow \,\,\,\,$$ y sinx = $$\int {4x\,dx\, + c} $$

$$ \Rightarrow \,\,\,\,$$ y sinx = 4.$${{{x^2}} \over 2} + c$$

$$ \Rightarrow \,\,\,\,$$ y sinx = 2x² + c . . . . . (1)

Given that, $$y\left( {{\pi \over 2}} \right) = 0$$

$$\therefore\,\,\,$$ x = $$y\left( {{\pi \over 2}} \right) = 0$$ and y = 0

Put this x = $${\pi \over 2}$$ and y = 0 at equation (1)

0.1 = 2. $$\left( {{\pi \over 2}} \right)$$² + c

$$ \Rightarrow \,\,\,$$ c =$$ - {{{\pi ^2}} \over 2}$$

So, differential equation is

y sin x = 2x² $$-$$ $${{{\pi ^2}} \over 2}\,\,.....(2)$$

Now we have to find y $$\left( {{\pi \over 6}} \right).$$

So, put x = $${{\pi \over 6}}$$ at equation (2)

y . sin $${{\pi \over 6}}$$ = 2 $${\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}$$

$$ \Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}$$ = 2. $${{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}$$

$$ \Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}$$

$$ \Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}$$

$$ \Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}$$