15．（5分）已知$f(x)=3\sin x+2$，对任意的$x_{1}\in [0$，$\dfrac{\pi }{2}]$，都存在$x_{2}\in [0$，$\dfrac{\pi }{2}]$，使得$f(x_{1})=2f(x_{2}+\theta )+2$成立，则下列选项中，$\theta$可能的值是(　　)
A．$\dfrac{3\pi }{5}$              B．$\dfrac{4\pi }{5}$              C．$\dfrac{6\pi }{5}$              D．$\dfrac{7\pi }{5}$
分析：由题意可知，$x_{1}\in [0$，$\dfrac{\pi }{2}]$，即$\sin x_{1}\in [0$，$1]$，可得$f(x_{1})\in [2$，$5]$，将存在任意的$x_{1}\in [0$，$\dfrac{\pi }{2}]$，都存在$x_{2}\in [0$，$\dfrac{\pi }{2}]$，使得$f(x)=2f(x+\theta )+2$成立，转化为$f(x_{2}+\theta )_{min}\leqslant 0$，$f({x}_{2}+\theta )_{max}\geqslant \dfrac{3}{2}$，又由$f(x)=3\sin x+2$，可得$\sin ({x}_{2}+\theta )_{min}\leqslant -\dfrac{2}{3}$，$\sin ({x}_{2}+\theta )_{max}\geqslant -\dfrac{1}{6}$，再将选项中的值，依次代入验证，即可求解．
解：$\because x_{1}\in [0$，$\dfrac{\pi }{2}]$，
$\therefore \sin x_{1}\in [0$，$1]$，
$\therefore f(x_{1})\in [2$，$5]$，
$\because$都存在$x_{2}\in [0$，$\dfrac{\pi }{2}]$，使得$f(x_{1})=2f(x_{2}+\theta )+2$成立，
$\therefore f(x_{2}+\theta )_{min}\leqslant 0$，$f({x}_{2}+\theta )_{max}\geqslant \dfrac{3}{2}$，
$\because f(x)=3\sin x+2$，
$\therefore$$\sin ({x}_{2}+\theta )_{min}\leqslant -\dfrac{2}{3}$，$\sin ({x}_{2}+\theta )_{max}\geqslant -\dfrac{1}{6}$，
$y=\sin x$在$x\in [\dfrac{\pi }{2},\dfrac{3\pi }{2}]$上单调递减，
当$\theta =\dfrac{3\pi }{5}$时，${x}_{2}+\theta \in [\dfrac{3\pi }{5},\dfrac{11\pi }{10}]$，
$\therefore$$\sin ({x}_{2}+\theta )=\sin \dfrac{11\pi }{10}>\sin \dfrac{7\pi }{6}=-\dfrac{1}{2}$，故$A$选项错误，
当$\theta =\dfrac{4\pi }{5}$时，${x}_{2}+\theta \in [\dfrac{4\pi }{5},\dfrac{13\pi }{10}]$，
$\therefore$$\sin ({x}_{2}+\theta )_{min}=\sin \dfrac{13\pi }{10}<\sin \dfrac{5\pi }{4}=-\dfrac{\sqrt{2}}{2}<-\dfrac{2}{3}$，
$\sin ({x}_{2}+\theta )_{max}=\sin \dfrac{4\pi }{5}>0$，故$B$选项正确，
当$\theta =\dfrac{6\pi }{5}$时，$x_{2}+\theta \in [\dfrac{6\pi }{5},\dfrac{17\pi }{10}]$，
$\sin (x_{2}+\theta )_{max}=\sin \dfrac{6\pi }{5}<\sin \dfrac{13\pi }{12}=\dfrac{\sqrt{2}-\sqrt{6}}{4}<-\dfrac{1}{6}$，故$C$选项错误，
当$\theta =\dfrac{7\pi }{5}$时，${x}_{2}+\theta \in [\dfrac{7\pi }{5},\dfrac{19\pi }{10}]$，
$\sin (x_{2}+\theta )_{max}=\sin \dfrac{19\pi }{10}<\sin \dfrac{23\pi }{12}=\dfrac{\sqrt{2}-\sqrt{6}}{4}<-\dfrac{1}{6}$，故$D$选项错误．
故选：$B$．
点评：本题考查了三角函数的单调性，以及恒成立问题，需要学生有较综合的知识，属于中档题．