How do you solve 2=4cos^2x+12=4cos2x+1 for 0<=x<=2pi0≤x≤2π?

1 Answer
sente
Dec 20, 2016

x in {pi/3, (2pi)/3, (4pi)/3, (5pi)/3}x∈{π3,2π3,4π3,5π3}

Explanation:

2=4cos^2(x)+12=4cos2(x)+1

=> 4cos^2(x)-1 = 0⇒4cos2(x)−1=0

=> cos^2(x)-1/4 = 0⇒cos2(x)−14=0

=> (cos(x)+1/2)(cos(x)-1/2) = 0⇒(cos(x)+12)(cos(x)−12)=0

=> cos(x)+1/2 = 0 or cos(x)-1/2 = 0⇒cos(x)+12=0orcos(x)−12=0

=> cos(x) = -1/2 or cos(x) = 1/2⇒cos(x)=−12orcos(x)=12

Checking a unit circle or relying on knowledge of common angles, we find that with the restriction x in [0, 2pi]x∈[0,2π], we have

cos(x) = -1/2 <=> x in {(2pi)/3, (4pi)/3}cos(x)=−12⇔x∈{2π3,4π3}
cos(x) = 1/2 <=> x in {pi/3, (5pi)/3}cos(x)=12⇔x∈{π3,5π3}

So, putting those together, we get our final result:

x in {pi/3, (2pi)/3, (4pi)/3, (5pi)/3}x∈{π3,2π3,4π3,5π3}

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