cos2x=cosx+sinx在[-2π,2π]的解集是
问题描述：
cos2x=cosx+sinx在[-2π,2π]的解集是
答
cos2x=cosx+sinx 两边平方 (cos2x)^2=(sinx+cosx)^2 (cos2x)^2=(sinx)^2+(cosx)^2+2sinxcosx 1-(sin2x)^2=1+sin2x (sin2x)^2+sin2x=0 sin2x(sin2x+1)=0 sin2x=0时 2x=2kπ x=kπ 取k=-2,-1,0,1,2 x=-2π,-π,0,π,2...