函数y=3cos((π/3)-2x)的递减区间是A.[kπ-(π/2),kπ+(5π/12)] (k∈z)B.[kπ+(5π/12),kπ+(11π/12)](k∈z)C.[kπ-(π/3),kπ+(π/6)](k∈z)D.[kπ+(π/6),kπ+(2π/3)](k∈z)
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函数y=3cos((π/3)-2x)的递减区间是A.[kπ-(π/2),kπ+(5π/12)](k∈z)B.[kπ+(5π/12),kπ+(11π/12)](k∈z)C.[kπ-(π/3),kπ+(π
函数y=3cos((π/3)-2x)的递减区间是A.[kπ-(π/2),kπ+(5π/12)] (k∈z)B.[kπ+(5π/12),kπ+(11π/12)](k∈z)C.[kπ-(π/3),kπ+(π/6)](k∈z)D.[kπ+(π/6),kπ+(2π/3)](k∈z)
函数y=3cos((π/3)-2x)的递减区间是
A.[kπ-(π/2),kπ+(5π/12)] (k∈z)
B.[kπ+(5π/12),kπ+(11π/12)](k∈z)
C.[kπ-(π/3),kπ+(π/6)](k∈z)
D.[kπ+(π/6),kπ+(2π/3)](k∈z)
函数y=3cos((π/3)-2x)的递减区间是A.[kπ-(π/2),kπ+(5π/12)] (k∈z)B.[kπ+(5π/12),kπ+(11π/12)](k∈z)C.[kπ-(π/3),kπ+(π/6)](k∈z)D.[kπ+(π/6),kπ+(2π/3)](k∈z)
选C啊
首先中括号里的代数式属于(0,π)闭区间,
然后解方程得.[-(π/3),+(π/6)]
因为周期kπ
所以选C