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已知函数f(x)=3cos^2x+2cosx*sinx+sin^2x
(1)求函数的周期
(2)写出f(x)的单调区间
(3)求f(x)的最大值,病求出此时x的值
人气:103 ℃ 时间:2020-06-05 13:50:46
解答
f(x)=3(cosx)^3+2sinxcosx+(sinx)^2
=sin2x+2(cosx)^2+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2

(1)最小正周期为T=2π/2=π,周期为kπ,k是不为0的整数.

(2)2kπ-π/2<2x+π/4<2kπ+π/2,则kπ-3π/82kπ+π/2<2x+π/4<2kπ+3π/2,则kπ+π/8
(3)当2x+π/4=2kπ+π/2,即x=kπ+π/8时,f(x)取得最大值为√2+2.
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