由cos²2x = 1 - 2sin²x,不等式可化为：
2sin²x+2sinx+2 > M - √(2M+1)
x∈R,于是 2sin²x+2sinx+2 = 2(sinx + 1/2)² + 3/2 ≥ 3/2
由题意,只需：M - √(2M+1) < 3/2
即：2M - 3 - 2√(2M+1) < 0
令 t = √(2M+1),则有：t² - 2t - 4 < 0
解得：0≤t
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