2. Given $\sin 2 x=\frac{\sin \theta+\cos \theta}{2}$, $\cos ^{2} x=\sin \theta \cdot \cos \theta$.
Then, the value of $\cos 2 x$ is ( ).
(A) $\frac{-1 \pm \sqrt{33}}{8}$
(B) $\frac{-1+\sqrt{33}}{8}$
(C) $\frac{-1-\sqrt{33}}{8}$
(D) 0
2. C.

Notice
$$
\begin{array}{l}
\sin ^{2} 2 x=\left(\frac{\sin \theta+\cos \theta}{2}\right)^{2}=\frac{1+2 \sin \theta \cdot \cos \theta}{4} \\
=\frac{1+2 \cos ^{2} x}{4},
\end{array}
$$

then $1-\cos ^{2} 2 x=\frac{2+\cos 2 x}{4}$.
Thus, $\cos 2 x=\frac{-1 \pm \sqrt{33}}{8}$.
$$
\begin{array}{l}
\text { Also, } \cos 2 x=2 \cos ^{2} x-1=2 \sin \theta \cdot \cos \theta-1 \\
=\sin 2 \theta-1<0,
\end{array}
$$

Therefore, $\cos 2 x=\frac{-1-\sqrt{33}}{8}$.