Let \(y=y(x)\) be the solution of the differential equation \(\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x, x \in(0, \pi)\).
If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to

\(\begin{aligned}
& \sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cos x=4 x \\
& \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}+y \cot x=\frac{4 x}{\sin x} \\
& \text { Here, } \mathrm{P}(x)=\cot x, \mathrm{Q}(x)=\frac{4 x}{\sin x} \\
& \text { Integrating factor (I.F.) }=\mathrm{e}^{\int \mathrm{P}(x) d x} \\
& =\mathrm{e}^{\int \operatorname{oot} x d x},=\mathrm{e}^{\log |\sin x|}=\sin x
\end{aligned}\)
\(\begin{array}{ll}
\therefore & y(\text { I.F. })=\int \mathrm{Q}(\text { I.F. }) \mathrm{d} x+\mathrm{c} \\
\therefore & y(\sin x)=\int \frac{4 x}{\sin x} \times \sin x \mathrm{~d} x=\mathrm{C} \\
\therefore & y \sin x=4 \int x \mathrm{~d} x+\mathrm{C} \\
\therefore & y=\frac{2 x^2+\mathrm{C}}{\sin x} \\
& y\left(\frac{\pi}{2}\right)=0
\end{array}\)...(i)
\(\begin{aligned} & \Rightarrow \frac{2\left(\frac{\pi}{2}\right)^2+C}{\sin \left(\frac{\pi}{2}\right)}=0 \\ & \Rightarrow \frac{\pi^2}{2}+C=0 \\ & \Rightarrow C=\frac{-\pi^2}{2}\end{aligned}\)
\(\Rightarrow y=\frac{2 x^2-\frac{\pi^2}{2}}{\sin x}\)
is the solution of the given differential equation.
\(\begin{aligned}
\therefore y\left(\frac{\pi}{6}\right) & =\frac{2\left(\frac{\pi}{6}\right)^2-\frac{\pi^2}{2}}{\sin \left(\frac{\pi}{6}\right)} \\
& =\frac{\frac{2 \pi^2}{36}-\frac{\pi^2}{2}}{\frac{1}{2}}=\frac{-8}{9} \pi^2
\end{aligned}\)