23. (8 points) Given $\sin \left(x+20^{\circ}\right)=\cos (x$ $+10^{\circ})+\cos \left(x-10^{\circ}\right)$. Find $\operatorname{tg} x$.
23. The given conditions can be transformed into
$$
\begin{array}{l}
\sin x \cos 20^{\circ}+\cos x \sin 20^{\circ}=2 \cos x \cos 10^{\circ}, \\
\sin x \cos 20^{\circ}=\cos x \cdot\left(2 \cos 10^{\circ}-\sin 20^{\circ}\right) . \\
\begin{aligned}
\therefore \operatorname{tg} x & =\frac{2 \cos 10^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}} \\
& =\frac{\cos 10^{\circ}+\sin 80^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}} \\
& =\frac{\cos 10^{\circ}+2 \cos 50^{\circ} \sin 30^{\circ}}{\cos 20^{\circ}} \\
& =\frac{2 \cos 30^{\circ} \cos 20^{\circ}}{\cos 20^{\circ}} \\
& =2 \cos 30^{\circ}=\sqrt{3} .
\end{aligned}
\end{array}
$$