The given equation is

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

Using the trigonometric identities $$\cos^2x = 1 - \sin^2x$$ and $$\cos^22x = 1 - 2\sin^2x$$, we can rewrite the equation in terms of $$\cos^2x$$:

$$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x$$

Simplify this equation :

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

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

We can factor out a 2 and rewrite this as :

$$\lambda = 2(\cos ^4 x - \cos ^2 x - \frac{1}{2})$$

$$\lambda = 2[(\cos ^2 x - \frac{1}{2})^2 - \frac{3}{4}]$$

So $\lambda_{\max }=2\left[\frac{1}{4}-\frac{3}{4}\right]=2 \times\left(\frac{-2}{4}\right)=-1$ (maximum value) and

$\lambda_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2}($ minimum value $)$

So range of the value of $x$ is $\left[\frac{-3}{2},-1\right]$.