- 难!三角函数的难题！非高手慎入～！！已知3sin^2a+2sin
- 已知3sin^2a+2sin^2b=2sina ,求c^2a+cos^2b 的取值范围 ～
- ∵3sin^2a+2sin^2b=2sina
∴整理得(sina+1/3)^2=-2/3sin^2b+1/9
∴ 0≤(sina+1/3)^2≤1/9
∴-2/3≤sina≤0
又由3sin^2a+2sin^2b=2sina得
sin^2a+2[2-(cos^2a+cos^2b)]=2sina
∴cos^2a+cos^2b=1/2(sina-1)^2+3/2
∵-2/3≤sina≤0
∴3/2≤cos^2a+cos^2b≤31/18
或许有错误，最好再算一下