How do you solve #2sin^2x=sinx#?

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Answer 1 · 2016-06-16T15:24:22

#x = {2kpi }uu {pi+2kpi }uu {pi/6+2kpi }uu {(5pi)/6+2kpi }#
for # {k=0,pm1,pm2,...}#

#2sin^2(x)=sin(x) -> sin (x)(2 sin (x)-1)=0#

so the conditions are

# {(sin(x) = 0), (2sin(x)-1=0) :}#

The solutions are

#x = {pi+2kpi }uu {2kpi }# for # {k=0,pm1,pm2,...}#

and

#x = {pi/6+2kpi }uu {(5pi)/6+2kpi }# for # {k=0,pm1,pm2,...}#

Finally

#x = {2kpi }uu {pi+2kpi }uu {pi/6+2kpi }uu {(5pi)/6+2kpi }#

for # {k=0,pm1,pm2,...}#