Let $y=y\left(x\right)$ be the solution of the differential equation  $\cos x\left(3\sin x+\cos x+3\right)dy=\left(1+y\sin x\left(3\sin x+\cos x+3\right)\right)dx,0\le x\le \frac{\pi }{2},y\left(0\right)=0$. Then, $y\left(\frac{\pi }{3}\right)$ is equal to:

$\begin{array}{l}\text{The given differential equation is }\\ cosx\left(3\sin x+\cos x+3\right)dy=\left(1+y\sin x\left(3\sin x+\cos x+3\right)\right)dx\\ \text{Rewrite the above equation as }\\ \frac{dy}{dx}=\frac{1+y\sin x\left(3\sin x+\cos x+3\right)}{\cos x\left(3\sin x+\cos x+3\right)}\\ =\frac{1}{\cos x\left(3\sin x+\cos x+3\right)}+y\tan x\\ \frac{dy}{dx}-y\tan x=\frac{1}{\cos x\left(3\sin x+\cos x+3\right)}\end{array}$

$$\begin{gathered} \text{This is a linear differential equation } \\ {\text{Integrating factor is e}}^{-\int tanxdx}=e^{-logsecx}=cosx \\ Hence, the solution of the given differenital equation is as below \end{gathered}$$

$\begin{array}{c}y\cos x=\int \frac{1}{\cos x\left(3\sin x+\cos x+3\right)}\cos xdx\\ =\int \frac{1}{\left(3\sin x+\cos x+3\right)}dx\end{array}$

$\text{Substitute }tan\frac{x}{2}=t,dx=\frac{2dt}{1+t^2},\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$

$\begin{array}{c}y\cos x=\int \frac{1}{3\cdot \frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}+3}\frac{2dt}{1+t^2}\\ =\int \frac{2}{6t+1-t^2+3+3t^2}dt\\ =\int \frac{2}{2t^2+6t+4}dt\\ =\int \frac{1}{t^2+3t+2}dt\\ =\int \frac{1}{\left(t+1\right)\left(t+2\right)}dt\\ =\int \frac{1}{t+1}-\frac{1}{t+2}dt\\ =\ln \left|t+1\right|-\ln \left|t+2\right|\\ =\ln \left|\frac{\tan \frac{x}{2}+1}{\tan \frac{x}{2}+2}\right|+c\end{array}$

$$\begin{gathered} \text{substitute x=0,y=0} \\ \text{it gives c=ln2} \end{gathered}$$

$\text{Therefore, the solution for the given differential equation is ycosx=ln}\left|\frac{1+tan{\displaystyle \frac{x}{2}}}{2+tan{\displaystyle \frac{x}{2}}}\right|$

$\text{when x=}\frac{\mathrm{\pi }}{3}$

$y\frac{1}{2}={\log }_e\left[\frac{1+\tan {\displaystyle \frac{\mathrm{\pi }}{6}}}{2+\tan {\displaystyle \frac{\mathrm{\pi }}{6}}}\right]+\ln \left(2\right)$

$$\begin{gathered} \text{simplifying the above expression } \\ {\text{we get y=2log}}_e\left[\frac{10+2\sqrt{3}}{11}\right] \end{gathered}$$