A function $$y = f(x)$$ satisfies $$f(x)\sin 2x + \sin x - (1 + \cos^2 x)f'(x) = 0$$ with condition $$f(0) = 0$$. Then $$f(\frac{\pi}{2})$$ is equal to

Given the ODE: $$f(x)\sin 2x + \sin x - (1 + \cos^2 x)f'(x) = 0$$ with $$f(0) = 0$$.

Rearrange the equation.

$$(1 + \cos^2 x)f'(x) - \sin 2x \cdot f(x) = \sin x$$

Recognize the exact derivative.

Note that $$\frac{d}{dx}[\cos^2 x] = -2\sin x\cos x = -\sin 2x$$. Therefore:

$$\frac{d}{dx}\left[(1 + \cos^2 x)f(x)\right] = (1 + \cos^2 x)f'(x) + (-\sin 2x)f(x)$$

This is exactly the left-hand side! So the equation becomes:

$$\frac{d}{dx}\left[(1 + \cos^2 x)f(x)\right] = \sin x$$

Integrate both sides.

$$(1 + \cos^2 x)f(x) = \int \sin x\,dx = -\cos x + C$$

Apply the initial condition $$f(0) = 0$$.

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

$$0 = -1 + C \implies C = 1$$

So the solution is:

$$(1 + \cos^2 x)f(x) = 1 - \cos x$$

Find $$f(\pi/2)$$.

$$(1 + \cos^2(\pi/2)) \cdot f(\pi/2) = 1 - \cos(\pi/2)$$

$$(1 + 0) \cdot f(\pi/2) = 1 - 0$$

$$f(\pi/2) = 1$$

The correct answer is Option (1): $$\boxed{1}$$.