5. Let $\pi \leqslant x<2 \pi$. Define
$$
\begin{aligned}
P= & \frac{1}{2} \cos x-\frac{1}{4} \sin 2 x-\frac{1}{8} \cos 3 x+ \\
& \frac{1}{16} \sin 4 x+\frac{1}{32} \cos 5 x-\frac{1}{64} \sin 6 x- \\
& \frac{1}{128} \cos 7 x+\cdots, \\
Q= & 1-\frac{1}{2} \sin x-\frac{1}{4} \cos 2 x+\frac{1}{8} \sin 3 x+ \\
& \frac{1}{16} \cos 4 x-\frac{1}{32} \sin 5 x-\frac{1}{64} \cos 6 x+ \\
& \frac{1}{128} \sin 7 x+\cdots .
\end{aligned}
$$

If $\frac{P}{Q}=\frac{2 \sqrt{2}}{7}$, then $\sin x=$ $\qquad$
5. $-\frac{17}{19}$.

From the problem, we have
$$
\begin{array}{l} 
Q+P \mathrm{i} \\
= 1+\frac{\mathrm{i}}{2}(\cos x+\mathrm{i} \sin x)+ \\
\frac{\mathrm{i}^{2}}{4}(\cos 2 x+\mathrm{i} \sin 2 x)+ \\
\frac{\mathrm{i}^{3}}{8}(\cos 3 x+\mathrm{i} \sin 3 x)+ \\
\frac{\mathrm{i}^{4}}{16}(\cos 4 x+\mathrm{i} \sin 4 x)+\cdots .
\end{array}
$$

Let $r=\frac{\mathrm{i}}{2} \mathrm{e}^{\mathrm{i} x}$. Then,
$$
Q+P \mathrm{i}=1+r+r^{2}+\cdots=\frac{1}{1-r} \text {. }
$$

By comparing the real and imaginary parts on both sides, we get
$$
\begin{array}{l}
P=\frac{2 \cos x}{5+2 \sin x}, Q=\frac{2 \sin x+4}{5+2 \sin x} \\
\Rightarrow \frac{P}{Q}=\frac{\cos x}{\sin x+2}=\frac{2 \sqrt{2}}{7} \\
\Rightarrow \sin x=-\frac{17}{19} .
\end{array}
$$