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已知函数fx=2cos²x+cos[2x+(π/3)]-1.(1)求函数fx的周期和单调递增区间.(2)若锐角a满足f(a)=-3/2,求a的值
人气:470 ℃ 时间:2020-02-03 14:49:19
解答
f(x)=1+cos2x+cos[2x+π/3]-1
=cos2x+cos[2x+π/3]
=2cos(2x+π/6)cos(π/6)
=√3cos(2x+π/6)
1)最小正周期T=2π/2=π
单调增区间:2kπ-π/2
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