Đáp án:
sin2x - sin2x cosx =0$$\\$$ =>sin2x.(1-cosx )=0$$\\$$ =>[(sin2x=0),(),(1-cosx =0):}$$\\$$ =>[(2x=kpi),(),(cosx =1):}$$\\$$ =>[(x=(kpi)/2),(),(x =k2pi):} (k in ZZ)$$\\$$ Vậy..
 

sin 2x - sin 2xcos x = 0
<=> sin 2x(1 - cos x) = 0
<=> [(sin 2x = 0),(1 - cos 2x = 0):}
<=> [(sin 2x = 0),(cos x = 1):}
<=> [(2x = k.\pi),(x = k.2\pi):} (k in ZZ)
<=> [(x = \frac{k.\pi}{2}),(x = k.2\pi):} (k in ZZ)
<=> x = \frac{k.\pi}{2} (k in ZZ)
Vậy ___________________________________________________