
1个回答
展开全部
f(x)=2cosx[1/2sinx+√3/2cosx]-√3sin^2(x)+sinxcosx
=sinxcosx+√3cos^2(x)-√3sin^2(x)+sinxcosx
=sin2x+√3cos2x
=2sin(2x+π/6)
当2x+π/6=π/2+2kπ时,即x=π/6+kπ时,f(x)取最大值, y(MAX) = 2
当2x+π/6=-π/2+2kπ时,即x= -π/3+kπ时,f(x)取最小值, y(min) = - 2
=sinxcosx+√3cos^2(x)-√3sin^2(x)+sinxcosx
=sin2x+√3cos2x
=2sin(2x+π/6)
当2x+π/6=π/2+2kπ时,即x=π/6+kπ时,f(x)取最大值, y(MAX) = 2
当2x+π/6=-π/2+2kπ时,即x= -π/3+kπ时,f(x)取最小值, y(min) = - 2
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询