If $$\frac{dy}{dx} + y\tan x = \sin 2x$$ and $$y(0) = 1$$, then $$y(\pi)$$ is equal to:

The given differential equation is:

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

with the initial condition $$y(0) = 1$$. This is a first-order linear differential equation of the form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

where $$P(x) = \tan x$$ and $$Q(x) = \sin 2x$$. To solve it, we use an integrating factor (IF). The integrating factor is defined as:

$$\text{IF} = e^{\int P(x) dx} = e^{\int \tan x dx}$$

First, compute $$\int \tan x dx$$:

$$\int \tan x dx = \int \frac{\sin x}{\cos x} dx$$

Let $$u = \cos x$$, then $$du = -\sin x dx$$, so:

$$\int \frac{\sin x}{\cos x} dx = -\int \frac{du}{u} = -\ln |u| + C = -\ln |\cos x| + C$$

For the integrating factor, we ignore the constant and take the absolute value appropriately. Thus:

$$\text{IF} = e^{-\ln |\cos x|} = e^{\ln \left( \frac{1}{|\cos x|} \right)} = \frac{1}{|\cos x|}$$

Since $$\cos x$$ can be positive or negative, but for simplicity in solving, we often use $$\sec x = \frac{1}{\cos x}$$ and handle the domain. In the interval around 0, $$\cos x > 0$$, so we can write:

$$\text{IF} = \sec x$$

Multiply both sides of the differential equation by the integrating factor:

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

The left-hand side is the derivative of $$y \sec x$$ because:

$$\frac{d}{dx} (y \sec x) = \sec x \frac{dy}{dx} + y \sec x \tan x$$

which matches the left-hand side. So:

$$\frac{d}{dx} (y \sec x) = \sec x \cdot \sin 2x$$

Simplify the right-hand side. Since $$\sin 2x = 2 \sin x \cos x$$,

$$\sec x \cdot \sin 2x = \frac{1}{\cos x} \cdot 2 \sin x \cos x = 2 \sin x$$

Thus:

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

Integrate both sides with respect to $$x$$:

$$\int \frac{d}{dx} (y \sec x) dx = \int 2 \sin x dx$$

$$y \sec x = 2 \int \sin x dx = 2 (-\cos x) + C = -2 \cos x + C$$

where $$C$$ is the constant of integration. Solve for $$y$$:

$$y \sec x = -2 \cos x + C$$

Multiply both sides by $$\cos x$$:

$$y = (-2 \cos x + C) \cos x = -2 \cos^2 x + C \cos x$$

So:

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

Use the initial condition $$y(0) = 1$$. When $$x = 0$$, $$\cos 0 = 1$$:

$$y(0) = C \cdot 1 - 2 \cdot (1)^2 = C - 2$$

Set equal to 1:

$$C - 2 = 1$$

$$C = 3$$

Substitute $$C = 3$$ into the solution:

$$y = 3 \cos x - 2 \cos^2 x$$

Now evaluate $$y(\pi)$$. When $$x = \pi$$, $$\cos \pi = -1$$:

$$y(\pi) = 3 \cdot (-1) - 2 \cdot (-1)^2 = -3 - 2 \cdot 1 = -3 - 2 = -5$$

Hence, the correct answer is Option D.