5. Let $\sin x+\cos x=\frac{1}{2}$, then $\sin ^{3} x+\cos ^{3} x=$
Solve $\frac{11}{16}$.
Analysis From $\sin x+\cos x=\frac{1}{2}$, we can get $1+2 \sin x \cos x=\frac{1}{4}$, so $\sin x \cos x=-\frac{3}{8}$, thus
$$
\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)\left(\sin ^{2} x-\cos x \sin x+\cos ^{2} x\right)=\frac{1}{2}\left(1+\frac{3}{8}\right)=\frac{11}{16}
$$