2. 2. 1 ^ Given the equation $\sin ^{2} x-(2 a+1) \cos x-a^{2}=0$ has real solutions for $x$. Then the set of real values for $a$ is ( ).
(A) $\left[-\frac{5}{4}, 1-\sqrt{2}\right]$
(B) $\left[-\frac{5}{4}, 1+\sqrt{2}\right]$
(C) $[1-\sqrt{2}, 1+\sqrt{2}]$
(D) $\left[-\frac{3}{2}, 1-\sqrt{2}\right]$
Transform the equation into $\cos ^{2} x+(2 a+1) \cos x+a^{2}-1=0$.
Let $t=\cos x$, then the equation becomes
$$
t^{2}+(2 a+1) t+a^{2}-1=0 .
$$

Let $f(t)=t^{2}+(2 a+1) t+a^{2}-1, t \in[-1,1]$. By the problem, the real number $a$ should satisfy
$$
\left\{\begin{array}{l}
(2 a+1)^{2}-4\left(a^{2}-1\right) \geqslant 0, \\
f(1) \geqslant 0, \\
f(-1) \geqslant 0, \\
-1 \leqslant-\frac{2 a+1}{2} \leqslant 1,
\end{array}\right.
$$

or $f(1) f(-1) \leqslant 0$. Solving this, we get $-\frac{5}{4} \leqslant a \leqslant 1+\sqrt{2}$.
Therefore, the set of values for the real number $a$ is $\left[-\frac{5}{4}, 1+\sqrt{2}\right]$, so the answer is B.