The number of values of $$x$$ in the interval $$\left[\frac{\pi}{4}, \frac{7\pi}{4}\right]$$ for which $$14\csc^2 x - 2\sin^2 x = 21 - 4\cos^2 x$$ holds, is ______.

Given,

$$14\csc^2x-2\sin^2x=21-4\cos^2x$$

Using

$$\csc^2x=\frac1{\sin^2x},\qquad \cos^2x=1-\sin^2x$$

we get

$$\frac{14}{\sin^2x}-2\sin^2x=21-4(1-\sin^2x)=17+4\sin^2x$$

$$\frac{14}{\sin^2x}=17+6\sin^2x$$

Let

$$t=\sin^2x$$

Then,

$$14=17t+6t^2$$

$$6t^2+17t-14=0$$

Factoring,

$$6t^2+21t-4t-14=0$$

$$(3t-2)(2t+7)=0$$

Thus,

$$t=\frac23\quad \text{or}\quad t=-\frac72$$

Since

$$\sin^2x\ge0$$

we reject

$$t=-\frac72$$

Hence,

$$\sin^2x=\frac23$$

$$\sin x=\pm\sqrt{\frac23}$$

Now,

$$\sqrt{\frac23}>\frac1{\sqrt2}$$

so, there are two solutions for

$$\sin x=\sqrt{\frac23}$$

in

$$\left[\frac\pi4,\frac{7\pi}4\right]$$

Also,

$$-\sqrt{\frac23}<-\frac1{\sqrt2}$$

so, there are two solutions for

$$\sin x=-\sqrt{\frac23}$$

in the same interval.

Therefore, the total number of solutions is $$\boxed{4}$$.