f(x)=2根号3sinxcosx+2cos^2x-1=√3sin2x+cos2x=2sin(2x+π/6)
kπ-π/3(cos2x)^2=16/25+2√3/5sin2x+3(sin2x)^2这里好像算错了-cos2x=4/5+√3sin2x ,两边平方(cos2x)^2=16/25+8√3/5sin2x+3(sin2x)^24(sin2x)^2+8√3/5sin2x-9/25=0sin2x=[-8√3/5+√(192/25+144/25)]/8=(-2√3+√21)/10