已知函数f（x）=2√3sinxcosx+2cosamp;#178;x-1,x∈R,1、求函数f（x）的最小正周期及在区间【0,π/2】上的最大值和最小值.2、若f（α）=6/5,α∈[π/4,π/2],求cos2α的值.
<p>问题：已知函数f（x）=2√3sinxcosx+2cosamp;#178;x-1,x∈R,1、求函数f（x）的最小正周期及在区间【0,π/2】上的最大值和最小值.2、若f（α）=6/5,α∈[π/4,π/2],求cos2α的值.<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">郭宇红的回答：<div class="content-b">网友采纳 1. f(x)=√3sin2x+cos2x =2sin(2x+π/6) (1) x∈ 2x+π/6∈[π/6,7π/6] 2x+π/6=π/2,x=π/6 f(x)min=f(7π/6)=-2sinπ/6=-1 f(x)max=f(π/6)=2sinπ/2=2 (2) sin(2α+π/6)=3/5 ∵α∈[π/4,π/2] 2α+π/6∈ cos(2α+π/6)=-4/5 cos2α=cos[(2α+π/6)-π/6]=(-4√3+3)/10<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">任姝婕的回答：<div class="content-b">网友采纳 1.f(x)min=f(7π/6)=-2sinπ/6=-1，为什么x=7π/6时有最小值？<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">郭宇红的回答：<div class="content-b">网友采纳 g(x)=sin(x)在[-π/2,π/2]单增,[π/2,3π/2]单减
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