f(x)=sin(2x+π/6)+sin(2x-π/6)-2cos²x,求值域及单调增区间
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f(x)=sin(2x+π/6)+sin(2x-π/6)-2cos²x,求值域及单调增区间
![f(x)=sin(2x+π/6)+sin(2x-π/6)-2cos²x,求值域及单调增区间](/uploads/image/z/17354660-68-0.jpg?t=f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2Bsin%282x-%CF%80%2F6%29-2cos%26%23178%3Bx%2C%E6%B1%82%E5%80%BC%E5%9F%9F%E5%8F%8A%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=sin(2x+π/6)+sin(2x-π/6)-2cos²x
=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x-2cos²x
=√3sin2x-1-cos2x
=2sin(2x-π/6)-1
∵sin(2x-π/6)∈[-1,1]
∴2sin(2x-π/6)-1∈[-3,1]
即:f(x)的值域是[-3,1]
单调增区间是:2kπ-π/2≤2x-π/6≤2kπ+π/2
解得:kπ-π/6≤x≤kπ+π/3
即:f(x)的单调增区间是[kπ-π/6,kπ+π/3] k∈Z
=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x-2cos²x
=√3sin2x-1-cos2x
=2sin(2x-π/6)-1
∵sin(2x-π/6)∈[-1,1]
∴2sin(2x-π/6)-1∈[-3,1]
即:f(x)的值域是[-3,1]
单调增区间是:2kπ-π/2≤2x-π/6≤2kπ+π/2
解得:kπ-π/6≤x≤kπ+π/3
即:f(x)的单调增区间是[kπ-π/6,kπ+π/3] k∈Z
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