3. The base of the pyramid $S-ABC$ is an equilateral triangle $ABC$, the side face $SAC$ is perpendicular to the base, and the dihedral angles formed by the other two side faces with the base are both $45^{\circ}$. Then the value of the dihedral angle $A$ $SC-B$ is $\qquad$ . (Express using inverse trigonometric functions)
3. $\operatorname{arctg} \sqrt{7}$.

Let $SO \perp AC$ at $O$, so $SO \perp$ plane $ABC$. Draw $SD \perp BC$ at $D$, $SE \perp AB$ at $E$, and connect $OD$, $OE$. Then $OD \perp BC$, $OE \perp AB$. According to the conditions, $\angle SDO = \angle SEO = 45^{\circ}$.
Thus, $SO - OE = OD$.
From the congruence of $\triangle SOF \simeq \triangle SOD$, $\triangle AOE \cong \triangle COD$. Therefore, $AO = CO = \frac{a}{2}$. ($a$ is the side length of the equilateral $\triangle ABC$) Draw $BM \perp SC$ at $M$, by the theorem of three perpendiculars, we get $OM \perp$

Thus, $\angle OMB$ is the plane angle of the dihedral angle $A-SC-B$. Calculation shows $SO = OD = \frac{\sqrt{3}}{4} a$, $SC = \frac{\sqrt{7}}{4} a$, $OM = \frac{OS \cdot OD}{SC} = \frac{\sqrt{3}}{2 \sqrt{7}} a$. Since $\operatorname{tg} \angle OMB = \frac{OB}{OM} = \sqrt{7}$, hence $\angle OMB = \operatorname{arctg} \sqrt{7}$.