Question

Question #984be

Answer 1

`sin2x=sin(x+pi/3)`

`=>sin2x-sin(x+pi/3)=0`

`=>2cos(3/2x+pi/6)sin(x/2-pi/6)=0`

When `cos(3/2x+pi/6)=0`

then `3/2x+pi/6=(2n+1)pi/2" where " n in ZZ`

`=>3/2x=npi+pi/2-pi/6`

`=>x=(2npi)/3+(2pi)/9=(2pi)/9(3n+1)`

when `sin(x/2-pi/6)=0`

then `(x/2-pi/6)=npi" where "n in ZZ`

`=>x=2npi+pi/3`