How do you solve #sin^2x - sinx =0# for #0<=x<=2pi#?

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Answer 1 · 2016-11-08T19:56:24

#{0,pi/2,pi, 2pi}#

#sin^2x-sinx=0#

Factorise

#sinx(sinx-1)=0#

solve for each factor.

#sinx =0#

#=>x=0, pi, 2pi#

or

#sinx-1=0=>sinx=1# for # 0<=x<=2pi#

only one solution #x=pi/2# for # 0<=x<=2pi#

solution set#{0,pi/2,pi, 2pi}#