sin^3x-cos^3x>cosx-sinx,求x的取值范围
问题描述:
sin^3x-cos^3x>cosx-sinx,求x的取值范围
x属于(0,2π)
答
sin^3x-cos^3x用立方差公式可得(sinx-cosx)(sin^2x+sinxcosx+cos^2x)
(sinx-cosx)(sin^2x+sinxcosx+cos^2x)>cosx-sinx
(sinx-cosx)(sin^2x+sinxcosx+cos^2x)-(sinx-conx)>0
(sinx-cosx)*sin2x>0
根号2*sin(x-π/4)*sin2x>0
跟着 你自己做吧