2.2. 2 * The maximum value of the real number $k$ for which the inequality $\frac{1+\sin x}{2+\cos x} \geqslant k$ has solutions for $x$ is $(\quad)$.
(A) $-\frac{1}{3}$
(B) $-\frac{3}{4}$
(C) $\frac{3}{4}$
(D) $\frac{4}{3}$
Given that $2+\cos x>0$, the original inequality can be transformed into $1+\sin x>k(2+\cos x)$, which is $\sin x - k \cos x>2 k-1$. Consequently, we have $\sqrt{1+k^{2}} \sin (x-\varphi)>2 k-1$. Solving the inequality $-\sqrt{1+k^{2}} \leqslant 2 k-1 \leqslant \sqrt{1+k^{2}}$, we find that the maximum value of $k$ is $\frac{4}{3}$, hence the answer is $\mathrm{D}$.