y'=cosx(1+cos)+sinx(0-sinx)=cosx+cos^2(x)-sin^2(x)=cosx+cos(2x)=2cos^2(x)+cosx-1cosx+2(cosx)^2-1=02(cosx)^2+cosx-1=0(cosx+1)(2cosx-1)=0cosx=-1或cosx=1/2x=π+2kπ或-π+2kπ或x=π/3+2kπ或-π/3+2kπ0≤x...x=π+2kπ或-π+2kπ或x=π/3+2kπ或-π/3+2kπk不是整数吗，k=0,1求出的x值在0≤x≤2π范围内只有这3个零点