How do you solve 2cos^3x + cos^2x = 02cos3x+cos2x=0 in the interval [0, 2pi]?

1 Answer
Hubert
Apr 25, 2016

x in {pi/2,(2pi)/3,(4pi)/3,(3pi)/2}x∈{π2,2π3,4π3,3π2}

Explanation:

2cos^3x+cos^2x=02cos3x+cos2x=0
2cos^2x(cosx+1/2)=02cos2x(cosx+12)=0
cos^2x=0 or cosx+1/2=0cos2x=0orcosx+12=0
cosx=0 or cosx=-1/2cosx=0orcosx=−12
x=pi/2+kpi or (x=(2pi)/3+2kpi or x=(4pi)/3+2kpi)x=π2+kπor(x=2π3+2kπorx=4π3+2kπ), where k in mathbb(Z)
Those are all real solutions but since we care about only the interval [0,2pi] the final solutions are: pi/2,(2pi)/3,(4pi)/3,(3pi)/2.

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