函数y=-2+2sin^2x化简.求最小正周期

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y=-2+2sin^2x
=-2+(1-cos2x)
=-cos2x-1
所以最小正周期为 2π/2=π
======以下答案可供参考======
供参考答案1:
解:y=-2+2sin^2x
=-2+1-cos2x
=-cos2x-1
最小正周期：T=2π/2=π