Question

高一数学

已知：3sin²α +2sin²β =2sinα ，求cos²α +cos²β 的取值范围

Answer 1

令 cos²α +cos²β =t
2cos²α +2cos²β =2t
3sin²α +2sin²β =2sinα 相加，得
sin^2a+4-2sina-2t=0
(sina-1)^2=2t-3
3sin²α +2sin²β =2sinα
2sinα-3sin²α =2sin²β>=0
2sinα-3sin²α>=0 0<=sina<=2/3
u=(sina-1)^2的范围是1/9<=u<=1
1/9<=2t-3<=1
14/9<=t<=2
cos²α +cos²β 的取值范围[14/9,2]

Answer 2

3sin²α +2sin²β =2sinα sin²β=sinα-(3/2)sin²α=1-cos²β cos²β=(3/2)sin²α-sinα+1
cos²α +cos²β =1-sin²α+(3/2)sin²α-sinα+1=(1/2)sin²α-sinα+2=(1/2)(sinα-1)^2+3/2
3/2<=cos²α +cos²β<=7/2

Answer 3