已知3sin2α+2sin2β=2sinα,求sin2α+sin2β的取值范围.
人气:457 ℃ 时间:2020-04-12 04:57:10
解答
∵3sin
2α+2sin
2β=2sinα,
∴sin
2β=sinα-
sin
2α≥0,
∴0≤sinα≤
;
∴sin
2α+sin
2β=sin
2α+sinα-
sin
2α
=
−(sinα−1)2+,
∵0≤sinα≤
;
∴
−(sinα−1)2+∈[0,].
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