f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
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f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间
![f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12)求单调递增区间](/uploads/image/z/18002711-47-1.jpg?t=f%28x%29%3D%E6%A0%B9%E5%8F%B73sin%282x-%CF%80%2F6%29%2B2sin%5E2%28x-%CF%80%2F12%29%E6%B1%82%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=√3sin(2x-π/6)+2sin²(x-π/12)
=√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2*sin(2x-π/6)-1/2*cos(2x-π/6)]+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3)+1
令2kπ-π/2≤2x-π/3≤2kπ+π/2,
得:kπ-π/12≤x≤kπ+5π/12
所以单调递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)
再问: 那递减区间呢
再答: 令2kπ+π/2≤2x-π/3≤2kπ+3π/2, 得:kπ+5π/12≤x≤kπ+11π/12 所以单调递增区间为[kπ+5π/12,kπ+11π/12] (k∈Z)
=√3sin(2x-π/6)+1-cos(2x-π/6)
=2[√3/2*sin(2x-π/6)-1/2*cos(2x-π/6)]+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3)+1
令2kπ-π/2≤2x-π/3≤2kπ+π/2,
得:kπ-π/12≤x≤kπ+5π/12
所以单调递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)
再问: 那递减区间呢
再答: 令2kπ+π/2≤2x-π/3≤2kπ+3π/2, 得:kπ+5π/12≤x≤kπ+11π/12 所以单调递增区间为[kπ+5π/12,kπ+11π/12] (k∈Z)
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