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求y=cos2x+2sinx-3的值域
人气:422 ℃ 时间:2020-04-10 15:56:41
解答
y=cos2x +2sinx -3
= 1-2(sinx)^2+2sinx -3
= -2(sinx-1/2)^2 - 3/2
max y = -3/2
min y = -2( -3/2)^2 - 5/2 = -9/2 -3/2 = -6
值域 = [-6,-3/2]
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