The set of all values of $\lambda$ for which the equation  ${\cos }^{2}2x-2{\sin }^{4}x-2{\cos }^{2}x=\lambda$ have solution is ....

The given equation is ${\cos }^{2}2x-2{\sin }^{4}x-2{\cos }^{2}x=\lambda$

Simplify the given equation in terms of $\cos 2x$

It implies that 

$\begin{array}{rcl}{\cos }^{2}2x-2{\left(\frac{1-\cos 2x}{2}\right)}^2-\left(1+\cos 2x\right)& =& \lambda \\ {\cos }^{2}2x-\frac{1}{2}\left(1+{\cos }^{2}2x-2\cos 2x\right)-1-\cos 2x& =& \lambda \\ {\cos }^{2}2x-\frac{1}{2}-\frac{1}{2}{\cos }^{2}2x+\cos 2x-1-\cos 2x& =& \lambda \\ \frac{1}{2}{\cos }^{2}2x-\frac{3}{2}-\lambda & =& 0\\ {\cos }^{2}2x-3-2\lambda & =& 0\end{array}$

The above equation can be written as 

$\begin{array}{rcl}{\cos }^{2}2x& =& 3+2\lambda \\ \cos 2x& =& \sqrt{3+2\lambda }\end{array}$

Hence,

$-1\le \sqrt{3+2\lambda }\le 1$

It gives

$3+2\lambda \le 1\Rightarrow 2\lambda \le -2\Rightarrow \lambda \le -1$

And

$3+2\lambda \ge 0\Rightarrow 2\lambda \ge -3\Rightarrow \lambda \ge -\frac{3}{2}$

Therefore, the required interval is $\left[-\frac{3}{2},-1\right]$