Let $ y = y(x) $ be the solution of the differential equation

$ x \frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,; x \ne 0 $.

If $ y(2) = x $, then $ \tan(y(1)) $ is equal to

$ x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y $

$ \sec^2 y \frac{dy}{dx} - 2\tan y \cdot \frac{1}{x} = x^2(2 - x^3) $

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

$ \frac{dt}{dx} - \frac{2t}{x} = x^2(2 - x^3) \quad (\text{LDE}) $

I.F. $ = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = \frac{1}{x^2} $

$ \therefore \frac{t}{x^2} = \int \frac{1}{x^2} x^2(2 - x^3)dx + C $

$ \frac{\tan y}{x^2} = 2x - \frac{x^4}{4} + C $

$ y(2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0 $

$ \tan y = 2x^3 - \frac{1}{4}x^6 $

$ x = 1 \Rightarrow \tan y = 2 - \frac{1}{4} = \frac{7}{4} $