Question

Question #79b74

Answer 1

`2cos2x-4cosx=1`

`=>2(2cos^2x-1)-4cosx=1`

`=>4cos^2x-2-4cosx=1`

`=>(2cosx)^2-2*cosx*1+1^2=1+2+1`

`=>(2cosx-1)^2=4`

`=>2cosx-1=pm2`

So

`cosx=(2+1)/2=1.5>1->"not possible"`

Again

`cosx=(-2+1)/2=-1/2=cos(pi-pi/3)`

`=>cosx=cos((2pi)/3)`

`=>x=2npipm(2pi)/3" where "n in ZZ`