Question

18．求函數(shù)y=lg（tanx-$\sqrt{3}$）+$\sqrt{2cosx+\sqrt{3}}$的定義域．

Answer 1

分析 根據(jù)函數(shù)y的解析式，列出使解析式有意義的不等式組$\left\{\begin{array}{l}{tanx-\sqrt{3}＞0}\\{2cosx+\sqrt{3}≥0}\end{array}\right.$，求出解集即可．

解答 解：∵函數(shù)y=lg（tanx-$\sqrt{3}$）+$\sqrt{2cosx+\sqrt{3}}$，
∴$\left\{\begin{array}{l}{tanx-\sqrt{3}＞0}\\{2cosx+\sqrt{3}≥0}\end{array}\right.$，
即$\left\{\begin{array}{l}{tanx＞\sqrt{3}}\\{cosx≥-\frac{\sqrt{3}}{2}}\end{array}\right.$；
解得$\left\{\begin{array}{l}{\frac{π}{3}+kπ＜x＜\frac{π}{2}+kπ，k∈Z}\\{-\frac{5π}{6}+2kπ≤x≤\frac{5π}{6}+2kπ，k∈Z}\end{array}\right.$，
即-$\frac{2π}{3}$+2kπ＜x＜-$\frac{π}{2}$+2kπ，或$\frac{π}{3}$+2kπ＜x＜$\frac{π}{2}$+2kπ，k∈Z；
∴該函數(shù)的定義域?yàn)閧x|-$\frac{2π}{3}$+2kπ＜x＜-$\frac{π}{2}$+2kπ，或$\frac{π}{3}$+2kπ＜x＜$\frac{π}{2}$+2kπ，k∈Z}．

點(diǎn)評(píng) 本題考查了求函數(shù)定義域的問(wèn)題，也考查了正切函數(shù)與余切函數(shù)的圖象與性質(zhì)的應(yīng)用問(wèn)題，是基礎(chǔ)題目．