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

We are given the differential equation

$$\cos x\,\frac{dy}{dx}+2y\sin x=\sin 2x,\qquad 0<x<\frac{\pi}{2}.$$

First we rewrite it in the standard linear form $$\frac{dy}{dx}+P(x)\,y=Q(x).$$ To do so, we divide every term by $$\cos x$$ (which is positive in the given interval):

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

Recalling $$\tan x=\dfrac{\sin x}{\cos x}$$ and $$\sin 2x=2\sin x\cos x,$$ we obtain

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

So the linear equation is

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

with $$P(x)=2\tan x$$ and $$Q(x)=2\sin x.$$

For a linear first-order ODE, the integrating factor is given by the formula

$$\text{I.F.}=e^{\int P(x)\,dx}.$$

We compute it step by step:

$$\int P(x)\,dx=\int 2\tan x\,dx=2\int\tan x\,dx=2\bigl(-\ln|\cos x|\bigr)=-2\ln(\cos x).$$

Hence

$$\text{I.F.}=e^{-2\ln(\cos x)}=(\cos x)^{-2}=\sec^{2}x.$$

Multiplying the whole differential equation by this integrating factor, we have

$$\sec^{2}x\frac{dy}{dx}+2\tan x\,\sec^{2}x\,y=2\sin x\,\sec^{2}x.$$

By construction of the integrating factor, the left side is the derivative of the product $$y\sec^{2}x$$. Indeed,

$$\frac{d}{dx}\bigl(y\sec^{2}x\bigr)=\sec^{2}x\frac{dy}{dx}+y\frac{d}{dx}\!\bigl(\sec^{2}x\bigr),$$

and since $$\frac{d}{dx}(\sec^{2}x)=2\tan x\sec^{2}x,$$ the two expressions match. Therefore

$$\frac{d}{dx}\bigl(y\sec^{2}x\bigr)=2\sin x\,\sec^{2}x.$$

We now integrate both sides with respect to $$x$$:

$$y\sec^{2}x=\int 2\sin x\,\sec^{2}x\,dx + C,$$

where $$C$$ is the constant of integration. To evaluate the integral, we write it out explicitly:

$$\int 2\sin x\,\sec^{2}x\,dx=\int 2\frac{\sin x}{\cos^{2}x}\,dx.$$

Let us set $$u=\cos x$$ so that $$du=-\sin x\,dx$$, or equivalently $$-\!du=\sin x\,dx$$. Substituting, we get

$$\int 2\frac{\sin x}{\cos^{2}x}\,dx = 2\int\frac{-du}{u^{2}} = -2\int u^{-2}\,du.$$

Because $$\int u^{-2}du = -u^{-1},$$ the integral becomes

$$-2\bigl(-u^{-1}\bigr)=2u^{-1}=\frac{2}{u}=\frac{2}{\cos x}=2\sec x.$$

Thus we have obtained

$$y\sec^{2}x = 2\sec x + C.$$

To isolate $$y$$ we multiply both sides by $$\cos^{2}x$$ (which equals $$1/\sec^{2}x$$):

$$y = 2\sec x\,\cos^{2}x + C\cos^{2}x.$$

Because $$\sec x\,\cos^{2}x=\dfrac{1}{\cos x}\cdot\cos^{2}x=\cos x,$$ this simplifies neatly:

$$y = 2\cos x + C\cos^{2}x.$$

Now we apply the initial condition $$y\!\left(\dfrac{\pi}{3}\right)=0.$$ We know $$\cos\!\left(\dfrac{\pi}{3}\right)=\dfrac12$$, so

$$0 = 2\cos\!\left(\frac{\pi}{3}\right) + C\cos^{2}\!\left(\frac{\pi}{3}\right) = 2\left(\frac12\right) + C\left(\frac12\right)^{2} = 1 + \frac{C}{4}.$$

Solving for $$C$$, we get

$$1+\frac{C}{4}=0 \;\Longrightarrow\; \frac{C}{4}=-1 \;\Longrightarrow\; C=-4.$$

Substituting this value back, the explicit solution becomes

$$y = 2\cos x - 4\cos^{2}x.$$

Finally we calculate $$y\!\left(\dfrac{\pi}{4}\right).$$ We recall $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}=\frac1{\sqrt2}$$ and hence $$\cos^{2}\!\left(\dfrac{\pi}{4}\right)=\frac12.$$ Therefore

$$\begin{aligned} y\!\left(\frac{\pi}{4}\right) &= 2\cos\!\left(\frac{\pi}{4}\right) - 4\cos^{2}\!\left(\frac{\pi}{4}\right)\\[4pt] &= 2\left(\frac{1}{\sqrt2}\right) - 4\left(\frac12\right)\\[4pt] &= \frac{2}{\sqrt2} - 2\\[4pt] &= \sqrt2 - 2. \end{aligned}$$

Hence, the correct answer is Option C.