Given,

${\cos }^{2}x·\frac{dy}{dx}-\left(\tan 2x\right)y={\cos }^{4}x$

$\Rightarrow \frac{dy}{dx}-\left(\frac{\tan 2x}{{\cos }^{2}x}\right)y=\frac{{\cos }^{4}x}{{\cos }^{2}x}$

$\Rightarrow \frac{dy}{dx}-\left(\frac{\tan 2x}{{\cos }^{2}x}\right)y={\cos }^{2}x$

This is the linear differential equation of the form $\frac{dy}{dx}+Py=Q$. Therefore, we get

$P=-\left(\frac{\tan 2x}{{\cos }^{2}x}\right)$ and $Q={\cos }^{2}x$

I.F. $=e^{\int Pdx}$

Now, 

$\int Pdx=-\int \left(\frac{\tan 2x}{{\cos }^{2}x}\right)dx$

             $=-\int \left(\frac{2\tan x}{1-{\tan }^{2}x}\right){\sec }^{2}xdx$

Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\int Pdx=\int \left(\frac{-2t}{1-t^2}\right)dt$

              $={\log }_e\left|1-t^2\right|$

$e^{\int Pdx}=e^{{\log }_e\left|1-{\tan }^{2}x\right|}=\left(1-{\tan }^{2}x\right)$

Now, the solution of the linear differential equation is

$y\times \mathrm{I}.\mathrm{F}.=\int \left(\mathrm{I}.\mathrm{F}.\times Q\right)dx+C$

$\Rightarrow y\left(1-{\tan }^{2}x\right)=\int \left\{\left(1-{\tan }^{2}x\right){\cos }^{2}x\right\}dx+C$

$\Rightarrow y\left(1-{\tan }^{2}x\right)=\int \left\{{\cos }^{2}x-{\sin }^{2}x\right\}dx+C$

$\Rightarrow y\left(1-{\tan }^{2}x\right)=\int \cos 2xdx+C$

$\Rightarrow y\left(1-{\tan }^{2}x\right)=\frac{\sin 2x}{2}+\frac{c}{2}$

$\Rightarrow y=\frac{1}{2}\left\{\frac{\sin 2x+c}{\left(1-{\tan }^{2}x\right)}\right\}$