Solve $-\cos \left(x+\frac{\pi}{6}\right)=\sin \left(x+\frac{\pi}{6}-\frac{\pi}{2}\right)$
$$=\sin \left(x-\frac{\pi}{3}\right),$$

The original equation can be transformed into
$$\begin{array}{l} \sin \left(x+\frac{\pi}{6}\right)=\sin \left(x-\frac{\pi}{3}\right) . \\ \therefore x+\frac{\pi}{6}=n \pi+(-1) \cdot\left(x-\frac{\pi}{3}\right) . \end{array}$$

When $\mathrm{B}=2 \mathrm{k}(\mathrm{k} \in \mathrm{Z})$, the equation becomes
$$x+\frac{\pi}{6}=2 k \pi+x-\frac{\pi}{3} \text {, no solution. }$$

When $n=2 k+1 \quad(k \in Z)$, the equation becomes
$$\begin{array}{c} x+\frac{\pi}{6}=(2 \mathrm{k}+1) \pi-x+\frac{\pi}{3}, \\ \mathbf{x}=\mathrm{k} \pi+\frac{7}{12} \pi . \end{array}$$

The solution set of the original equation is $\left\{\mathbf{x}: \mathbf{x}=\mathrm{k} \pi+\frac{7}{12} \pi, \mathrm{k} \in \mathrm{Z}\right\}$.