If $$y = y(x)$$ is the solution of the differential equation, $$\frac{dy}{dx} + 2y\tan x = \sin x$$, $$y\left(\frac{\pi}{3}\right) = 0$$, then the maximum value of the function $$y(x)$$ over $$R$$ is equal to:

The differential equation is $$\frac{dy}{dx} + 2y\tan x = \sin x$$, which is a first-order linear ODE. The integrating factor is $$e^{\int 2\tan x\,dx} = e^{-2\ln|\cos x|} = \sec^2 x$$.

Multiplying both sides by $$\sec^2 x$$: $$\frac{d}{dx}(y\sec^2 x) = \sin x \cdot \sec^2 x = \frac{\sin x}{\cos^2 x} = \sec x \tan x$$.

Integrating: $$y\sec^2 x = \int \sec x \tan x\,dx = \sec x + C$$. So $$y = \cos x + C\cos^2 x$$.

Applying the initial condition $$y(\pi/3) = 0$$: $$0 = \cos(\pi/3) + C\cos^2(\pi/3) = \frac{1}{2} + C \cdot \frac{1}{4}$$, giving $$C = -2$$.

Therefore $$y = \cos x - 2\cos^2 x$$. To find the maximum, let $$u = \cos x$$, so $$y = u - 2u^2$$, where $$u \in [-1, 1]$$. Setting $$\frac{dy}{du} = 1 - 4u = 0$$ gives $$u = \frac{1}{4}$$, and the maximum value is $$y = \frac{1}{4} - 2 \cdot \frac{1}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$$.