How do you solve 3cos2x+cosx=-13cos2x+cosx=−1 for 0<=x<=2pi0≤x≤2π?

1 Answer
Binayaka C.
Sep 19, 2016

Solution: x=30^0,131.81^0,228.19^0,300^0x=300,131.810,228.190,3000 for 0<=x<=2pi0≤x≤2π

Explanation:

3cos 2x+cosx = -1 or 3(2cos^2x-1)+cosx +1=0 or 6cos^2x+cosx -2=03cos2x+cosx=−1or3(2cos2x−1)+cosx+1=0or6cos2x+cosx−2=0. Since we know cos2x=2cos^2x-1cos2x=2cos2x−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]

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