(sinx)^2+((√3)/2)sin2x-2(cosx)^2-m=0在x属于[0,π/2]上有实数解,则求m的取值范

1、由公式cos2x=2(cosx)^2 -1
即(cosx)^2=0.5cos2x +0.5可知,
f(x)=√3/2 sin2x -0.5cos2x -1
=sin(2x-π/6) -1
x属于[-π/12,5π/12],所以2x-π/6属于[-π/3,2π/3]
显然当2x-π/6=π/2,即x=π/3时,f(x)取最大值,f(π/3)=sin(π/2)-1=0
当2x-π/6= -π/3,即x= -π/12时,f(x)取最小值,f(-π/12)=sin(-π/3)-1= -0.5√3 -1
2、f(C)=0,即sin(2C-π/6)=1,
所以2C-π/6=π/2,解得C=π/3,
而向量m=(1,sinA)与向量n=(2,sinB)共线,
即sinB/sinA=2,
A+B+C=π,C=π/3
所以B=2π/3 -A
故sin(2π/3 -A) /sinA=2,
而sin(2π/3 -A)=sin(2π/3)cosA -cos(2π/3)sinA=0.5√3cosA +0.5sinA
于是(0.5√3cosA +0.5sinA) /sinA
= 0.5√3cotA+0.5=2
解得cotA=√3,即A=π/6,
故B=2π/3 -A=π/2,
所以a=c×tanA=√3×tan(π/6)=1,
b=a/ sinA=1/sin(π/6)=2

由题得：f(x)=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1
所以,当2x-π/6=2kπ-π/2 时,(k∈Z)
即：当,x=kπ-π/6 时,(k∈Z)
f(x)有最小值=-2,最小正周期=2π/2=π

f（x）=√3/2sin2x-cos²x-1/2
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)-1
f(C)=0
则C=π/3
因为sinB=2sinA
根据正弦定理b=2a .1
cosC=（a²+b²-c²）/（2ab）=1/2 .2
由1,2得
a=1
b=2