一直喊是Y=2sinx方+sin2x,x属于[0,2π]求是f(x)为正值的x集合

题目

一直喊是Y=2sinx方+sin2x,x属于[0,2π]求是f(x)为正值的x集合

答案

1 cos2x=cosxcosx-sinxsinx=1-2sinxsinx 得2sinxsinx=1-cos2x 2 f(x)=1-cos2x+sin2x=1+[根2(sin2xcosπ/4-cos2xsinπ/4)] =1+根2sin(2x-π/4) 3 f(x)>0 即 1+根2sin(2x-π/4)>0 sin(2x-π/4)>-根2/2 得 -π/4+2kπ...