Question

Question #00441

Answer 1

`cos(x - pi/6)=(sqrt3)cosx`

`=>cosx cos(pi/6+sinxsin(pi/6)=(sqrt3)cosx`

`=>cosx (sqrt3/2)+sinx(1/2)=(sqrt3)cosx`

`=> sqrt3cosx+sinx=(2sqrt3)cosx`

`=> sinx=(2sqrt3)cosx-sqrt3cosx`

`=> sinx=sqrt3cosx`

`=> sinx/cosx=sqrt3`

`=> tanx=tan(pi/3)`

`=>x= npi+pi/3" where " n in ZZ`