已知函数F(x)=2根号3sinxcosx+cos²x-sin²x-1,求单调递增区间;

若x∈[-5π/12,π/3]求f(x)取值范围

化简可得 f(x)=2sin(2x+Pi/6)-1
单调递增区间:-Pi/2<2x+Pi/6<Pi/2
-Pi/3<x<Pi/6
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第1个回答  2013-08-30
f(x)=根号3sin2x+cos2x-1=2sin(2x+π/6)-1
2kπ-π/2<2x+π/6<2kπ+π/2,所以单调增区间为[kπ-π/3,kπ+π/6]
f(x)属于[-3,1]
思路就是这样。
但是不知道算的过程中有没有错

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