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已知f(x)=sin(2x+л/3)+sin(2x-л/3) g(x)=根号三cos2x (1)设

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已知f(x)=sin(2x+л/3)+sin(2x-л/3) g(x)=根号三cos2x (1)设
已知f(x)=sin(2x+л/3)+sin(2x-л/3) g(x)=根号三cos2x
(1)设h(x)=f(x)g(x)求h(x)单调递增区间
f(x)=sin(2x+π3)+sin(2x-π/3)
f(x)=sin(2x)cos(π/3)+co(2x)sin(π/3)+sin(2x)cos(π/3)-cos(2x)sin(π/3)
f(x)=2sin(2x)cos(π/3)
f(x)=√3sin(2x)
g(x)=√3cos2x
则:
h(x)=f(x)g(x)=3sin(2x)cos(2x)
h(x)=(3/2)sin(4x)
函数h(x)的递增区间是:
2kπ-π/2≤4x≤2kπ+π/2
得:
(1/2)kπ-(π/8)≤x≤(1/2)kπ+(π/8)
递增区间是:[(1/2)kπ-(π/8),(1/2)kπ+(π/8)],其中k∈Z