Question #06395

Question

1094 views around the world

You can reuse this answer
Creative Commons License

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

$sin2x+sinx=0$

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

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

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