Question

How do you solve `3cos2x+cosx=-1` for `0<=x<=2pi`?

Answer 1

Solution: `x=30^0,131.81^0,228.19^0,300^0` for `0<=x<=2pi`

Explanation:

`3cos 2x+cosx = -1 or 3(2cos^2x-1)+cosx +1=0 or 6cos^2x+cosx -2=0`. Since we know `cos2x=2cos^2x-1`
`6cos^2x+cosx -2=0 or (2cosx-1)(3cosx+2)=0 :. 2cosx=1 :. cosx=1/2 : cos60=1/2 , cos 300=1/2:. x=60^0, x=300^0` OR `3cosx=-2 or cosx= -2/3 :. x=cos^-1(-2/3)=131.81^0 ;`Also `cos((180-131)+180)= -2/3 :. x=228.19^0`.So solution: `x=30^0,131.81^0,228.19^0,300^0` for `0<=x<=2pi`[Ans]