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

$\begin{array}{l}\frac{dy}{dx}-\frac{\tan 2x}{{\cos }^{2}x}y={\cos }^{2}x\\ \text{ I.E. }={\mathrm{e}}^{-\int \frac{\tan 2\mathrm{x}}{{\cos }^{2}x}\mathrm{dx}}\mathrm{dx}=e^{-\int \frac{2\tan x}{1-{\tan }^{2}x}se{c}^{2}x dx}\\ put \tan x=t, se{c}^{2}x dx=dt\\ =e^{\int \frac{-2t}{1-t^2}dt}=e^{\log (1-t^2)}=1-{\tan }^{2}x\\ \text{ Solution of D.E is }\mathrm{y}(1-{\tan }^2\mathrm{x})=\int (1-{\tan }^2\mathrm{x}){\cos }^{2}x\mathrm{dx}\\ y\left(1-\frac{{\sin }^{2}x}{{\cos }^{2}x}\right)=\int \left(1-\frac{{\sin }^{2}x}{{\cos }^{2}x}\right){\cos }^{2}x dx\\ \frac{y\cos 2x}{{\cos }^{2}x}=\int \cos 2x dx\\ \frac{y\cos 2x}{{\cos }^{2}x}=\frac{\sin 2x}{2}+C\\ \text{ When }x=\pi /6,y=\frac{3\sqrt{3}}{8}\\ \frac{\frac{3\sqrt{3}}{8}\cos {\displaystyle \frac{\pi }{3}}}{{\cos }^2{\displaystyle \frac{\pi }{6}}}=\frac{\sin {\displaystyle \frac{\pi }{3}}}{2}+C\\ \frac{3\sqrt{3}}{8}\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3}{4}}}=\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{2}+C\\ \Rightarrow C=0\\ \frac{y\cos 2x}{{\cos }^{2}x}=\frac{\sin 2x}{2}\\ \mathrm{y}=\frac{1}{2}\tan 2{\mathrm{xcos}}^2\mathrm{x}\end{array}$