复合三角函数的单调增区间

解:
y
=2√3sinxcosx+2cos^2(x)-1
=√3(2sinxcosx)+2*[1+cos2x]/2-1
=√3sin2x+cos2x
=2sin(2x+π/6)
(辅助角公式)

设T=2x+π/6
则:2sin(2x+π/6)的单调增区间为:
2kπ-π/2<=T<=2kπ+π/2
即:2kπ-π/2=<(2x+π/6)<=2kπ+π/2
则:kπ-π/3=<x<=kπ+π/6
则:
函数的单调增区间为:
[kπ-π/3,kπ+π/6]

y=√3sin2x+cos2x+2=2sin(2x+π/6)+2=2sin[2(x+π/12)]+2
然后画图就行了

=√3sin2x+cos2x+2
=2sin(2x+60du)
下面我就不写了，你自己会算的

y=2√3sinxcosx+2cos平方x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
=2sin[2(x+π/12)]
2kπ-π/2<=2(x+π/12)<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6