Let \(y=y(x)\) be the solution of the differential equation \(x\frac{dy}{dx}−\sin 2y={x}^{3}\left(2−{x}^{3}\right){\cos …

\(x\frac{dy}{dx}−\sin 2y={x}^{3}\left(2−{x}^{3}\right){\cos }^{2}y\)

\({\sec }^{2}y\frac{dy}{dx}−2\tan y⋅\frac{1}{x}={x}^{2}\left(2−{x}^{3}\right)\)

\(\tan y=t\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{dt}{dx}\)

\(\frac{dt}{dx}−\frac{2t}{x}={x}^{2}\left(2−{x}^{3}\right)\)(LDE)

I.F. \(={e}^{\int −\frac{2}{x}dx}={e}^{−2\ln x}=\frac{1}{{x}^{2}}\)

\(∴\frac{t}{{x}^{2}}=\int \frac{1}{{x}^{2}}{x}^{2}\left(2−{x}^{3}\right)dx+C\)

\(\frac{\tan y}{{x}^{2}}=2x−\frac{{x}^{4}}{4}+C\)

\(y\left(2\right)=0\Rightarrow 0=4−4+C\Rightarrow C=0\)

\(\tan y=2{x}^{3}−\frac{1}{4}{x}^{6}\)

at \(x=1\Rightarrow \tan y=2−\frac{1}{4}=\frac{7}{4}\)