已知方程（sinx+cosx）^2/(2+2sin2x-(cos2x)^2)

1,f(x)的定义域就是分母不等于0
2+2sin2x-(cos2x)^2
=2+2sin2x-[1-(sin2x)^2]
=(sin2x)^2+2sin2x+1
=(sin2x+1)^2≠0
sin2x≠-1
x≠2kπ-π/2
值域:sin2x+1=(sinx+cosx)^2
所以f(x)=1/(sinx+cosx)^2=1/(sin2x+1)
所以f(x)≥1/2

2，f(x)=2
1/(sin2x+1)=2
sin2x=-1/2
因为-π/4<x<3π/4
所以-π/2<2x<3π/2
所以2x=-π/6或5π/6或7π/6
所以x=-π/12或5π/12或7π/12

老兄，貌似题目不全啊
方程不是方程啊