Question #06395

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Answer 1 · 2017-03-08T10:07:05

#sin2x+sinx=0#

#=2sin((3x)/2)cos(x/2)=0#

When

#sin((3x)/2)=0#

#=>(3x)/2=npi"where " n in ZZ#

#=>x=(2npi)/3"where " n in ZZ#

When

#cos((x/2)=0#

#=>x/2=(2n+1)pi/2"where " n in ZZ#

#=>x=(2n+1)pi"where " n in ZZ#

Alternative

#sin2x+sinx=0#

#sin2x=-sinx=sin(-x)#

#=>2x=npi-(-1)^nx" where "n in ZZ#

#=>x=(npi)/(2+(-1)^n)" where "n in ZZ#