1）f(x)=2sinxcosx-cos2x=sin2x-cos2x=√2sin(2x-π/4)fmax=√2fmin=-√22）f(x)=√2sin(2x-π/4)=1sin(2x-π/4)=√2/22x-π/4=2kπ+π/4或2kπ+3π/4x=kπ+π/4或kπ+π/2又x∈[0,2π]∴x=π/4,π/2,5π/4,3π/2则 ...∴x=0,π/4，π，5π/4这里是怎么个写法上面写错一步，我改过来了~算出 x=kπ+π/4或kπ+π/2令k=0和k=1算出x=π/4或x=π/2或x=5π/4或x=3π/2分别带入求解。