求sin2x-2sinx-cosx+1=0的通解.

∵sin2x-2sinx-cosx+1=0 ==>2sinxcosx-2sinx-cosx+1=0
==>2sinx(cosx-1)-(cosx-1)=0
==>(2sinx-1)(cosx-1)=0
==>2sinx-1=0,或cosx-1=0
==>sinx=1/2,或cosx=1
∴x=π/6,或x=5π/6,或x=0
故原方程的通解是x=2kπ+π/6,或x=2kπ+5π/6,或x=2kπ (k是整数).