Let $$y=y(x)$$ be the solution of the differential equation $$2\cos x\,\frac{dy}{dx}= \sin 2x - 4y\sin x,x\in\left(0,\frac{\pi}{2}\right).$$ If $$y\!\left(\frac{\pi}{3}\right)=0$$, then $$y'\!\left(\frac{\pi}{4}\right)+ y\!\left(\frac{\pi}{4}\right)$$ is equal to $$\underline{\hspace{2cm}}.$$

$$2\cos x \frac{dy}{dx} = \sin 2x - 4y\sin x$$, $$x \in (0, \pi/2)$$.

$$2\cos x \frac{dy}{dx} + 4y\sin x = 2\sin x\cos x$$

$$\frac{dy}{dx} + 2y\tan x = \sin x$$

This is linear with integrating factor $$e^{\int 2\tan x\,dx} = e^{-2\ln\cos x} = \sec^2 x$$.

$$\frac{d}{dx}(y\sec^2 x) = \sin x \sec^2 x = \frac{\sin x}{\cos^2 x} = \sec x \tan x$$

Integrating: $$y\sec^2 x = \sec x + C$$, so $$y = \cos x + C\cos^2 x$$.

Using $$y(\pi/3) = 0$$: $$0 = \cos(\pi/3) + C\cos^2(\pi/3) = 1/2 + C/4$$.

$$C = -2$$.

$$y = \cos x - 2\cos^2 x$$.

$$y' = -\sin x + 4\sin x\cos x = -\sin x + 2\sin 2x$$.

At $$x = \pi/4$$:

$$y(\pi/4) = \frac{1}{\sqrt{2}} - 2 \times \frac{1}{2} = \frac{1}{\sqrt{2}} - 1$$

$$y'(\pi/4) = -\frac{1}{\sqrt{2}} + 4 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} + 2$$

$$y'(\pi/4) + y(\pi/4) = \left(-\frac{1}{\sqrt{2}} + 2\right) + \left(\frac{1}{\sqrt{2}} - 1\right) = 1$$.