f(x)=2sinxcosx+2√3(cosx)^2-√3+2
=sin2x+√3*cos2x+2
=2sin(2x+π/3)+2
因為x∈(0,π/2)
所以2x+π/3∈(π/3,4π/3)
所以2sin(4π/3)+2＜f(x)≤2sin(π/2)+2
即2-√3＜f(x)≤4
因為函數(shù)g(x)=f(x)+m有零點
所以m的取值范圍是(-4,√3-2]
如果不懂,請Hi我,祝學習愉快!
補充：第二問
f(x)=2sin(2x+π/3)+2
f(x0)=2sin(2x0+π/3)+2=2/5
所以sin(2x0+π/3)=-4/5
x0∈(π/4,π/2)
2x0+π/3∈(5π/6,4π/3)(結合它的正弦是負的,那么它是第三象限角)
故cos(2x0+π/3)=-√[1-(-4/5)^2]=-3/5
所以sin(2x0)=sin[(2x0+π/3)-π/3]=sin(2x0+π/3)cos(π/3)-cos(2x0+π/3)sin(π/3)=(-4/5)*(1/2)-(-3/5)*(√3/2)=(3√3-4)/10