How do you solve 2cos^2x+cosx=02cos2x+cosx=0 and find all exact general solutions?

2 Answers
Dec 16, 2016

General Solution for 2cos^2x+cosx=02cos2x+cosx=0 is

x=((2n+1)pi)/2x=(2n+1)π2 or x=2npi+-(2pi)/3x=2nπ±2π3, where nn is an integer.

Explanation:

2cos^2x+cosx=02cos2x+cosx=0

hArrcosx(2cosx+1)=0cosx(2cosx+1)=0

i.e. either cosx=0cosx=0 or 2cosx+1=02cosx+1=0 i.e. cosx=-1/2cosx=12

General solution for cosx=0cosx=0 is x=((2n+1)pi)/2x=(2n+1)π2, where nn is an integer

and general solution for cosx=-1/2=cos(+-(2pi)/3)cosx=12=cos(±2π3) is x=2npi+-(2pi)/3x=2nπ±2π3, where nn is an integer

Hence, General Solution for 2cos^2x+cosx=02cos2x+cosx=0 is

x=((2n+1)pi)/2x=(2n+1)π2 or x=2npi+-(2pi)/3x=2nπ±2π3, where nn is an integer.

Dec 16, 2016

The solutions are S={pi/2+2pin,3/2pi+2pin,2/3pi+2pin,4/3pi+2pin} S={π2+2πn,32π+2πn,23π+2πn,43π+2πn}

Explanation:

The equation is

2cos^2+cosx=02cos2+cosx=0

cosx(2cosx+1)=0cosx(2cosx+1)=0

cosx=0cosx=0 and cosx=-1/2cosx=12

cosx=0cosx=0 gives x=pi/2+2pinx=π2+2πn and x=3/2pi+2pinx=32π+2πn

cosx=-1/2cosx=12 gives x=2/3pi+2pinx=23π+2πn and x=4/3pi+2pinx=43π+2πn