Question

Question #52f58

Answer 1

`cos^2 3x+3sin3x=3`

`=>1-sin^2 3x+3sin3x=3`

`=>sin^2 3x-3sin3x+2=0`

`=>sin^2 3x-2sin3x-sin3x+2=0`

`=>sin3x(sin3x-2)-(sin3x-2)=0`

`=>(sin3x-2)(sin3x-1)=0`

`sin3x=2->"not possible"`

So `sin3x=1=sin(pi/2)`

`=>3x=npi+(-1)^npi/2" where "ninZZ`

`=>x=(npi)/3+(-1)^npi/6" where "ninZZ`