Answer
\[y(x)=\sin 2x+C\cos x\]
Work Step by Step
$2(\cos^2 x)y'+y\sin 2x=4\cos^4 x$ ____(1)
$2(\cos^2 x)\frac{dy}{dx}+2y\sin x\cos x=4\cos^4 x$
$\frac{dy}{dx}+y\tan x=2\cos^2 x$ ____(2)
This is linear differential equation
Integrating factor:-
\[I(x)=e^{\int\tan xdx}=e^{\ln|\sec x|}=\sec x\]
Multiply (2) by $I(x)$
$\sec x\frac{dy}{dx}+y\sec x\tan x=2 \cos x$
$\frac{d}{dx}[y\sec x]=2\cos x$
Integrating,
$y\sec x=2\int\cos x dx+C$
$C$ is of constant of integration
$y\sec x=2\sin x+C$
$y=2\sin x\cos x+C\cos x$
$y(x)=\sin 2x+C\cos x$
Hence general solution of (1) is $y(x)=\sin 2x+C\cos x$
