(2010•东城区模拟)函数f(x)=sin2x−3cos2x
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(2010•东城区模拟)函数f(x)=sin2x−
cos2x
3 |
![(2010•东城区模拟)函数f(x)=sin2x−3cos2x](/uploads/image/z/8226338-50-8.jpg?t=%EF%BC%882010%E2%80%A2%E4%B8%9C%E5%9F%8E%E5%8C%BA%E6%A8%A1%E6%8B%9F%EF%BC%89%E5%87%BD%E6%95%B0f%28x%29%EF%BC%9Dsin2x%E2%88%923cos2x)
sin2x-
3cos2x
=2(
1
2sin2x-
3
2cos2x)
=2(cos
π
3sin2x-sin
π
3cos2x)
=2sin(2x-
π
3)
∴y=2sin(2x-
π
3)
∵2kπ-
π
2≤当2x-
π
3≤2kπ+
π
2(k∈Z),即kπ−
π
12≤x≤kπ+
5π
12(k∈Z)时,
函数y=2sin(2x-
π
3)单调递增.
∴函数f(x)=sin2x−
3cos2x的单调递增区间为[−
π
12+kπ,
5π
12+kπ],k∈Z
故答案为:[−
π
12+kπ,
5π
12+kπ],k∈Z
3cos2x
=2(
1
2sin2x-
3
2cos2x)
=2(cos
π
3sin2x-sin
π
3cos2x)
=2sin(2x-
π
3)
∴y=2sin(2x-
π
3)
∵2kπ-
π
2≤当2x-
π
3≤2kπ+
π
2(k∈Z),即kπ−
π
12≤x≤kπ+
5π
12(k∈Z)时,
函数y=2sin(2x-
π
3)单调递增.
∴函数f(x)=sin2x−
3cos2x的单调递增区间为[−
π
12+kπ,
5π
12+kπ],k∈Z
故答案为:[−
π
12+kπ,
5π
12+kπ],k∈Z
(2010•东城区模拟)函数f(x)=sin2x−3cos2x
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