Given that, sin2x-sinx=0. rArr 2sinxcosx-sinx=0. rArr sinx(2cosx-1)=0. rArr sinx=0, or, cosx=1/2sin2x−sinx=0.⇒2sinxcosx−sinx=0.⇒sinx(2cosx−1)=0.⇒sinx=0,or,cosx=12
The zeros of sin are, kpi, k in ZZ. Since we require zeros in [0,2pi], these are 0,pi,2pi.
Next cosx=1/2=cos(pi/3) rArr x=2kpi+-pi/3, k in ZZ.This is since that the general soln. of the eqn. costheta=cosalpha is theta=2kpi+-alpha, k in ZZ.
k=0, rArr x=+pi/3 in [0,2pi], as -pi/3!in[0,2pi]
k=1 rArr x=2pi-pi/3=5pi/3, as 2pi+pi/3 !in [0,2pi].
Altogether, the soln. set ={0,pi/3,pi,5pi/3,2pi}.