The number of solutions of the equation $$\sin 2x - 2\cos x + 4\sin x = 4$$ in the interval $$[0, 5\pi]$$ is :

We start with the equation $$\sin 2x - 2\cos x + 4\sin x = 4$$. We need to find the number of solutions in the interval $$[0, 5\pi]$$.

First, recall that $$\sin 2x = 2 \sin x \cos x$$. Substitute this into the equation:

$$2 \sin x \cos x - 2\cos x + 4\sin x = 4$$

Move all terms to the left side to set the equation to zero:

$$2 \sin x \cos x - 2\cos x + 4\sin x - 4 = 0$$

Factor out the common factor of 2:

$$2(\sin x \cos x - \cos x + 2\sin x - 2) = 0$$

Divide both sides by 2:

$$\sin x \cos x - \cos x + 2\sin x - 2 = 0$$

Group the terms to factor by grouping. Group the first two terms and the last two terms:

$$(\sin x \cos x - \cos x) + (2\sin x - 2) = 0$$

Factor out $$\cos x$$ from the first group and 2 from the second group:

$$\cos x (\sin x - 1) + 2(\sin x - 1) = 0$$

Notice that $$(\sin x - 1)$$ is a common factor. Factor it out:

$$(\sin x - 1)(\cos x + 2) = 0$$

This product is zero when either factor is zero. So, we have two cases:

Case 1: $$\sin x - 1 = 0$$

$$\sin x = 1$$

Case 2: $$\cos x + 2 = 0$$

$$\cos x = -2$$

Now, solve each case separately.

For Case 1: $$\sin x = 1$$. The solutions occur at $$x = \frac{\pi}{2} + 2k\pi$$, where $$k$$ is an integer.

For Case 2: $$\cos x = -2$$. Since the cosine function ranges between -1 and 1, $$\cos x = -2$$ is impossible. So, there are no solutions for this case.

Thus, only Case 1 gives solutions: $$x = \frac{\pi}{2} + 2k\pi$$.

Now, find all solutions in the interval $$[0, 5\pi]$$. Substitute $$x = \frac{\pi}{2} + 2k\pi$$ and solve for integer $$k$$ such that:

$$0 \leq \frac{\pi}{2} + 2k\pi \leq 5\pi$$

Divide all terms by $$\pi$$:

$$0 \leq \frac{1}{2} + 2k \leq 5$$

Subtract $$\frac{1}{2}$$ from all parts:

$$-\frac{1}{2} \leq 2k \leq \frac{9}{2}$$

Divide by 2:

$$-\frac{1}{4} \leq k \leq \frac{9}{4}$$

Since $$k$$ must be an integer, the possible values are $$k = 0, 1, 2$$.

Now, find the corresponding $$x$$ values:

For $$k = 0$$: $$x = \frac{\pi}{2} + 2(0)\pi = \frac{\pi}{2}$$

For $$k = 1$$: $$x = \frac{\pi}{2} + 2(1)\pi = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$$

For $$k = 2$$: $$x = \frac{\pi}{2} + 2(2)\pi = \frac{\pi}{2} + 4\pi = \frac{9\pi}{2}$$

Check if these are within $$[0, 5\pi]$$:

$$\frac{\pi}{2} \approx 1.57$$, which is between 0 and $$5\pi \approx 15.7$$

$$\frac{5\pi}{2} = 2.5\pi \approx 7.85$$, which is between 0 and 15.7

$$\frac{9\pi}{2} = 4.5\pi \approx 14.137$$, which is less than $$5\pi \approx 15.7$$, so it is included

Now, check $$k = 3$$: $$x = \frac{\pi}{2} + 2(3)\pi = \frac{\pi}{2} + 6\pi = \frac{13\pi}{2} = 6.5\pi \approx 20.42$$, which is greater than $$5\pi \approx 15.7$$, so not in the interval

Check $$k = -1$$: $$x = \frac{\pi}{2} + 2(-1)\pi = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2} \approx -4.71$$, which is less than 0, so not in the interval

Therefore, the solutions are $$x = \frac{\pi}{2}$$, $$x = \frac{5\pi}{2}$$, and $$x = \frac{9\pi}{2}$$.

Verify each solution in the original equation:

At $$x = \frac{\pi}{2}$$:

$$\sin 2\left(\frac{\pi}{2}\right) = \sin \pi = 0$$

$$-2 \cos \left(\frac{\pi}{2}\right) = -2(0) = 0$$

$$+4 \sin \left(\frac{\pi}{2}\right) = 4(1) = 4$$

Sum: $$0 + 0 + 4 = 4$$, which satisfies the equation

At $$x = \frac{5\pi}{2}$$: same as $$x = \frac{\pi}{2}$$ because the functions are periodic with period $$2\pi$$, so it satisfies

At $$x = \frac{9\pi}{2}$$: same as $$x = \frac{\pi}{2}$$, so it satisfies

No other solutions exist because $$\cos x = -2$$ is impossible, and $$\sin x = 1$$ only occurs at these points in the interval.

Thus, there are three solutions in $$[0, 5\pi]$$.

Hence, the correct answer is Option A.