Question

已知函数f(x)=2√3sinxcosx-2cos2x+2，求：（1）f(x)的单调递增区间

急急急，求帮忙

Answer 1

F（X）=根号3sin2x cos2x +1
= 2sin（2X-π/ 6）+1
ORDER2kπ-π/ 2 <2X-π/ 6 <2kπ+π/ 2 />Kπ-π/ 6 <州轿桥X <Kπ+π/ 3
增册猛加帆慧间隔[Kπ-π/ 6Kπ+π/ 3]

Answer 2

F（X）=根举兄迅号3sin2x的cos2x +1

= 2sin（2X-π/ 6）尘如，+ 1的
ORDER2kπ-π/ 2 <2X-π/ 6 2 />Kπ-π/ 6 <X <Kπ+π/ 3
增加间隔[Kπ-π/正此6Kπ的+π/ 3]

Answer 3

2根号3sinxcosx=根号3sin2x
√√√√√√√√√√√√√√√√√