How do you solve cos^2x= cosxcos2x=cosx for x in the interval [0,2pi)?

1 Answer
sente
Mar 1, 2016

Move all terms to one side and factor the equation to find all possible values for cos(x)cos(x) to find that the solution set to cos^2(x) = cos(x)cos2(x)=cos(x) restricted to [0, 2pi)[0,2π) is

{0, pi/2, (3pi)/2}{0,π2,3π2}

Explanation:

cos^2(x) = cos(x)cos2(x)=cos(x)

=> cos^2(x) - cos(x) = 0⇒cos2(x)−cos(x)=0

=> cos(x)(cos(x)-1) = 0⇒cos(x)(cos(x)−1)=0

=> cos(x) = 0⇒cos(x)=0 or cos(x) = 1cos(x)=1

On the interval [0, 2pi)[0,2π) we have

cos(x) = 0 <=> x in {pi/2, (3pi)/2}cos(x)=0⇔x∈{π2,3π2}

and

cos(x) = 1 <=> x = 0cos(x)=1⇔x=0

Thus our complete solution set is x in {0, pi/2, (3pi)/2}x∈{0,π2,3π2}

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