2. Point $P\left(\ln \left(2^{x}+2^{-x}-\tan \frac{\pi}{6}\right), \cos 2\right)(x$ $\in \mathbf{R})$ is located in the ( ).
(A) first quadrant
(B) second quadrant
(C) third quadrant
(D) fourth quadrant
2.D.

Since $2^{x}+2^{-x}-\tan \frac{\pi}{6} \geqslant 2-\frac{\sqrt{3}}{3}>1$, therefore,
$$
\begin{array}{l}
\ln \left(2^{x}+2^{-x}-\tan \frac{\pi}{6}\right)>0 . \\
\text { Also } \frac{\pi}{2}<2<\pi, \text { so } \cos 2<0 .
\end{array}
$$