949

已知 ${\displaystyle \alpha ,\beta }$ 是函数 ${\displaystyle f(x)={\displaystyle 3\sin (2x+\frac{\pi }{6})-2}}$ 在 ${\displaystyle {\displaystyle (0,\frac{\pi }{2})}}$ 上的两个零点，求 ${\displaystyle \cos (\alpha -\beta )}$.

由题意及和差化积公式$$\begin{array}{r}0=\sin (2\alpha +\frac{\pi }{6})-\sin (2\beta +\frac{\pi }{6})=2\cos (\alpha +\beta +\frac{\pi }{6})\sin (\alpha -\beta ),\end{array}$$
题目的"两个零点"指 ${\displaystyle \alpha \ne \beta }$, 因此只保留 ${\displaystyle \cos {\displaystyle (\alpha +\beta +\frac{\pi }{6})=0}}$, 进而由 ${\displaystyle {\displaystyle \alpha +\beta +\frac{\pi }{6}\in (\frac{\pi }{6},\frac{7\pi }{6})}}$ 知 ${\displaystyle {\displaystyle \sin (\alpha +\beta +\frac{\pi }{6})=1}}$. 进而$$\begin{array}{rl}\cos (\alpha -\beta )=& \cos ((\alpha +\beta +\frac{\pi }{6})-(2\beta +\frac{\pi }{6}))\\ =& \cos (\alpha +\beta +\frac{\pi }{6})\cos (2\beta +\frac{\pi }{6})+\sin (\alpha +\beta +\frac{\pi }{6})\sin (2\beta +\frac{\pi }{6})\\ =& 1\cdot \sin (2\beta +\frac{\pi }{6})=\frac{2}{3}.\end{array}$$