$5.16 \cos ^{3} x+\cos ^{2} x-4 \cos ^{2} \frac{x}{2}=0$.
5.16 By replacing $2 \cos ^{2} \frac{x}{2}$ with $1+\cos x$ and factoring the left side of the equation, we get

$$
\begin{aligned}
& \cos ^{3} x+\cos ^{2} x-2(1+\cos x)=0 \\
& \cos ^{2} x(\cos x+1)-2(1+\cos x)=0 ;(1+\cos x)\left(\cos ^{2} x-2\right)=0 \\
& \cos ^{2} x-2 \neq 0, \text { since }|\cos x| \leqslant 1 ; 1+\cos x=0 ; \cos x=-1 \\
& x=\pi+2 \pi k
\end{aligned}
$$

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