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