The general solution of the differential equation, $$\sin 2x\left(\frac{dy}{dx} - \sqrt{\tan x}\right) - y = 0$$, is:

The given differential equation is:

$$\sin 2x\left(\frac{dy}{dx} - \sqrt{\tan x}\right) - y = 0$$

First, expand the equation:

$$\sin 2x \cdot \frac{dy}{dx} - \sin 2x \cdot \sqrt{\tan x} - y = 0$$

Recall that $$\sin 2x = 2 \sin x \cos x$$, so substitute:

$$2 \sin x \cos x \cdot \frac{dy}{dx} - 2 \sin x \cos x \cdot \sqrt{\tan x} - y = 0$$

Rearrange the terms to isolate the derivative:

$$2 \sin x \cos x \cdot \frac{dy}{dx} - y = 2 \sin x \cos x \cdot \sqrt{\tan x}$$

Divide both sides by $$\sin x \cos x$$:

$$\frac{2 \sin x \cos x \cdot \frac{dy}{dx}}{\sin x \cos x} - \frac{y}{\sin x \cos x} = \frac{2 \sin x \cos x \cdot \sqrt{\tan x}}{\sin x \cos x}$$

Simplify:

$$2 \frac{dy}{dx} - \frac{y}{\sin x \cos x} = 2 \sqrt{\tan x}$$

Note that $$\frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$$, so:

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

Divide both sides by 2:

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

This is a linear differential equation of the form $$\frac{dy}{dx} + P(x) y = Q(x)$$, where $$P(x) = -\frac{1}{\sin 2x}$$ and $$Q(x) = \sqrt{\tan x}$$.

The integrating factor (I.F.) is $$e^{\int P(x) dx} = e^{\int -\frac{1}{\sin 2x} dx}$$.

Compute $$\int -\frac{1}{\sin 2x} dx$$. Since $$\frac{1}{\sin 2x} = \csc 2x$$, we have $$\int -\csc 2x dx$$.

Let $$\theta = 2x$$, so $$d\theta = 2 dx$$ and $$dx = \frac{d\theta}{2}$$:

$$\int -\csc \theta \cdot \frac{d\theta}{2} = -\frac{1}{2} \int \csc \theta d\theta$$

The integral of $$\csc \theta$$ is $$\ln |\csc \theta - \cot \theta| + C$$, so:

$$-\frac{1}{2} \int \csc \theta d\theta = -\frac{1}{2} \ln |\csc \theta - \cot \theta| + C$$

Substitute back $$\theta = 2x$$:

$$-\frac{1}{2} \ln |\csc 2x - \cot 2x| + C$$

Simplify $$\csc 2x - \cot 2x$$:

$$\csc 2x - \cot 2x = \frac{1}{\sin 2x} - \frac{\cos 2x}{\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$$

Use identities $$1 - \cos 2x = 2 \sin^2 x$$ and $$\sin 2x = 2 \sin x \cos x$$:

$$\frac{2 \sin^2 x}{2 \sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$$

Thus:

$$\int -\frac{1}{\sin 2x} dx = -\frac{1}{2} \ln |\tan x| + C$$

The integrating factor is:

$$I.F. = e^{-\frac{1}{2} \ln |\tan x|} = e^{\ln |\tan x|^{-1/2}} = |\tan x|^{-1/2}$$

Since $$\sqrt{\tan x}$$ is in the equation, we consider $$\tan x \gt 0$$, so drop the absolute value:

$$I.F. = (\tan x)^{-1/2} = \frac{1}{\sqrt{\tan x}} = \sqrt{\cot x}$$

The solution of the linear differential equation is:

$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$

$$y \sqrt{\cot x} = \int \sqrt{\tan x} \cdot \sqrt{\cot x} dx + C$$

Simplify the integrand:

$$\sqrt{\tan x} \cdot \sqrt{\cot x} = \sqrt{\tan x \cdot \cot x} = \sqrt{1} = 1$$

Thus:

$$y \sqrt{\cot x} = \int 1 dx + C = x + C$$

Comparing with the options:

A. $$y\sqrt{\tan x} = x + c$$

B. $$y\sqrt{\cot x} = \tan x + c$$

C. $$y\sqrt{\tan x} = \cot x + c$$

D. $$y\sqrt{\cot x} = x + c$$

Hence, the correct answer is Option D.