4. Calculate: $\sqrt{9+8 \cos 20^{\circ}}-\sec 20^{\circ}=$
4.3.

From the triple angle formula, we get
$8 \cos ^{3} 20^{\circ}=1+6 \cos 20^{\circ}$.
Then $\sqrt{9+8 \cos 20^{\circ}}-\sec 20^{\circ}$
$$
=\frac{\sqrt{9 \cos ^{2} 20^{\circ}+8 \cos ^{3} 20^{\circ}}-1}{\cos 20^{\circ}}=3 .
$$