3. For what values of the parameter a does the equation $\cos ^{4} 2 x-2(a+2) \cos ^{2} 2 x-(2 a+5)=0$ have at least one solution
Solution: Let's make the substitution $\cos ^{2} 2 x=t, t \in[0 ; 1]$, then the equation will take the form: $t^{2}-2(a+2) t-(2 a+5)=0$. We will solve the quadratic equation with respect to $t$: $t_{1,2}=\frac{2(a+2) \pm \sqrt{4(a+2)^{2}+4(2 a+5)}}{2}=\frac{2 a+4 \pm 2(a+3)}{2} ; t_{1}=-1, t_{2}=2 a+5$.

Notice that $t=-1 \notin[0 ; 1]$.

Then, for the equation to have at least one solution, the condition $t_{2} \in[0 ; 1]$ must be satisfied, that is $0 \leq 2 a+5 \leq 1 \Leftrightarrow-2.5 \leq a \leq-2$.

Answer: $\mathrm{a} \in[-2.5 ;-2]$.