5.55 Since $\operatorname{ctg}^{4} x \geqslant 0, \cos ^{2} 2 x-1<0$, we arrive at the system of equations

$\left\{\begin{array}{l}\operatorname{ctg}^{4} x=0 \\ \cos ^{2} 2 x-1=0 .\end{array}\right.$

From this, we find: 1) $\left.x=\frac{\pi}{2}(2 k+1) ; 2\right) x=\frac{\pi k}{2}$, but in this case, for $k=21$, $\operatorname{ctg} x$ does not exist. Therefore, $x=\frac{\pi}{2}(2 k+1)$.

Answer: $\quad x=\frac{\pi}{2}(2 k+1), k \in Z$.