The set of all values of $$\lambda$$ for which the equation $$\cos^2 2x - 2\sin^4 x - 2\cos^2 x = \lambda$$

We need to find the range of $$\lambda$$ for which $$\cos^2 2x - 2\sin^4 x - 2\cos^2 x = \lambda$$ has solutions.

To begin,

Using $$\cos 2x = 1 - 2\sin^2 x$$:

$$\cos^2 2x = (1 - 2\sin^2 x)^2 = 1 - 4\sin^2 x + 4\sin^4 x$$

Substituting:

$$\lambda = 1 - 4\sin^2 x + 4\sin^4 x - 2\sin^4 x - 2\cos^2 x$$

Using $$\cos^2 x = 1 - \sin^2 x$$:

$$\lambda = 1 - 4\sin^2 x + 2\sin^4 x - 2(1 - \sin^2 x)$$

$$= 1 - 4\sin^2 x + 2\sin^4 x - 2 + 2\sin^2 x$$

$$= 2\sin^4 x - 2\sin^2 x - 1$$

Next,

Let $$t = \sin^2 x$$, where $$t \in [0, 1]$$.

$$\lambda = f(t) = 2t^2 - 2t - 1$$

From this,

$$f'(t) = 4t - 2 = 0 \implies t = \frac{1}{2}$$

Evaluating at critical point and endpoints:

$$f(0) = -1$$

$$f\left(\frac{1}{2}\right) = 2 \times \frac{1}{4} - 2 \times \frac{1}{2} - 1 = \frac{1}{2} - 1 - 1 = -\frac{3}{2}$$

$$f(1) = 2 - 2 - 1 = -1$$

Minimum value = $$-\frac{3}{2}$$, Maximum value = $$-1$$.

Therefore, $$\lambda \in \left[-\frac{3}{2}, -1\right]$$.

The correct answer is Option 4: $$\left[-\frac{3}{2}, -1\right]$$.