5. The roots of the equation $2 x^{2}+p x+q=0$ are $\sin \alpha$ and $\cos \alpha$. Then, in the $p o q$ coordinate plane, the figure of the point $(p, q)$ is ( ).


5. (B).

From the relationship between roots and coefficients, we have
$$
\left\{\begin{array}{l}
-\frac{p}{2}=\sin \alpha+\cos \alpha, \\
\frac{q}{2}=\sin 2 \alpha .
\end{array}\right.
$$

Substituting (2) into (1) gives $\frac{p^{2}}{4}=1+q$. Therefore, $q=\frac{1}{4} p^{2}-1$.
Since $|\sin \alpha+\cos \alpha| \leqslant \sqrt{2}$, from (1) we know $-2 \sqrt{2} \leqslant p \leqslant 2 \sqrt{2}$. From (2), $q=2 \sin \alpha \cos \alpha=\sin 2 \alpha$.
Thus, $-1 \leqslant q \leqslant 1$.
Therefore, the answer is (B).