已知函数f(x)=sinxcosx+√3cos^2x-√3/2,求最小正周期,求在[0,3π)内使f(x)取得最大值的所有x的和
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T=π; 符合条件所有x和为:3π+π/4
f(x)=sinxcosx+√3cos^2x-√3/2,
=1/2 sin2x + √3 (1+cos2x)/2 --√3/2
=1/2 sin2x +√3/2 cos2x
=sin(2x+π/3)
所以,T=2π/w=2π/2=π
f(x)取得最大值,即f(x)=sin(2x+π/3)=1,即2x+π/3=2kπ + π/2,
求得,x=kπ+π/12,在[0,3π)内使f(x)取得最大值的x有,
k=0, x=π/12; k=1, x=π + π/12;k=2, x=2π + π/12
所以,所有x的和为为:π/12 + π+π/12 + 2π+π/12=3π + π/4
f(x)=sinxcosx+√3cos^2x-√3/2,
=1/2 sin2x + √3 (1+cos2x)/2 --√3/2
=1/2 sin2x +√3/2 cos2x
=sin(2x+π/3)
所以,T=2π/w=2π/2=π
f(x)取得最大值,即f(x)=sin(2x+π/3)=1,即2x+π/3=2kπ + π/2,
求得,x=kπ+π/12,在[0,3π)内使f(x)取得最大值的x有,
k=0, x=π/12; k=1, x=π + π/12;k=2, x=2π + π/12
所以,所有x的和为为:π/12 + π+π/12 + 2π+π/12=3π + π/4
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