1、已知2根号3sin^2α+(2根号2-根号3)sinαcosα-根号2cos^2α=0,α∈(π/2,π),求sin

1、
2√3sin^2 α+(2√2-√3)sinαcosα-√2cos^2 α=0
===> (2sinα-cosα)(√3sinα+√2cosα)=0
===> 2sinα=cosα（不可能!因为在第二象限,sinα与cosα异号）或者,√3sinα+√2cosα=0
===> sinα=√10/5,cosα=-√15/5
所以,sin2α=2sinαcosα=-2√6/5
2、
f(x)=4sinxsin^2 x(π/4+x/2)+cos2x
=4sinx*[(1-cos(π/2+x))/2]+cos2x
=2sinx(1+sinx)+1-2sin^2 x
=2sinx+2sin^2 x+1-2sin^2 x
=2sinx+1
所以,f(ωx)=2sinωx+1
则,周期T=2π/ω=π
所以,ω=2
则,f(ωx)=f(2x)=2sin2x+1
g(x)=f^2(2x)+2f(2x)=(2sin2x+1)^2+2(2sin2x+1)
=4sin^2 2x+4sin2x+1+4sin2x+2
=4sin^2 2x+8sin2x+3
=4(sin2x+1)^2-1
因为sin2x∈[-1,1],则sin2x+1∈[0,2]
所以,g(x)的最大值为4*4-1=15
5、
cos2B=cosB
===> 2cos^2 B-1=cosB
===> 2cos^2 B-cosB-1=0
===> (2cosB+1)(cosB-1)=0
===> cosB=-1/2,或者cosB=1
===> B=2π/3,或者B=0（舍去）
已知B=2π/3,所以：A+C=π/3
则,sinA+sinC=sinA+sin(π/3-A)
=sinA+sin(π/3)cosA-cos(π/3)sinA
=sinA+(√3/2)cosA-(1/2)sinA
=(1/2)sinA+(√3/2)cosA
=sin[A+(π/3)]
因为A∈(0,π/3),所以A+(π/3)∈(π/3,2π/3)
所以,sinA+sinC=sin[A+(π/3)]∈(√3/2,1)