Solution of the equation ${\cos }^2\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-(\tan 2\mathrm{x})\mathrm{y}={\cos }^4\mathrm{x},\left|\mathrm{x}\right|<\frac{\mathrm{\pi }}{4}$, when $\mathrm{y}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{3\sqrt{3}}{8}$ is

The given differential equation can be written as $\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\tan 2\mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{y}={\cos }^2\mathrm{x}$ which is linear differential equation of first order.
$\begin{array}{l}\int \mathrm{Pdx}=\int \frac{-\sin 2\mathrm{x}}{\cos 2{\mathrm{xcos}}^2\mathrm{x}}\mathrm{dx}\\ =-\int \frac{2\sin 2\mathrm{xdx}}{\cos 2\mathrm{x}(1+\cos 2\mathrm{x})}\\ =\int \frac{\mathrm{dt}}{\mathrm{t}(1+\mathrm{t})}\\ =\int \left(\frac{1}{\mathrm{t}}-\frac{1}{1+\mathrm{t}}\right)\mathrm{dt}\\ =\log \frac{\mathrm{t}}{1+\mathrm{t}}\text{ where }\mathrm{t}=\cos 2\mathrm{x}\\ =\log \frac{\cos 2\mathrm{x}}{1+\cos 2\mathrm{x}}\left[\therefore -\frac{\mathrm{\pi }}{2}<2\mathrm{x}<\frac{\mathrm{\pi }}{2}\right]\end{array}$