1. The maximum value of $y=\sin \left(\frac{\pi}{3}+x\right)-\sin 3 x$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) 1
(C) $\frac{16 \sqrt{3}}{25}$
(D) $\frac{8 \sqrt{3}}{9}$
1.D.
$$
\begin{array}{l}
\text { Let } \frac{\pi}{3}+x=t \text {, then } x=t-\frac{\pi}{3}, \\
\sin 3 x=\sin (3 t-\pi)=-\sin 3 t, \\
y=\sin \left(\frac{\pi}{3}+x\right)-\sin 3 x=\sin t+\sin 3 t \\
=4 \sin t\left(1-\sin ^{2} t\right) .
\end{array}
$$

Thus, $y^{2}=16 \sin ^{2} t\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right)$
$$
\begin{array}{l}
=8\left(2 \sin ^{2} t\right)\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right) \\
\leqslant 8 \times\left(\frac{2}{3}\right)^{3}=\frac{64}{27} .
\end{array}
$$

Therefore, $y \leqslant \frac{8 \sqrt{3}}{9}$, equality holds when $\sin ^{2} t=\frac{1}{3}$.