How do you solve $2=4cos^2x+1$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-12-20T01:11:04

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

$2=4cos^2(x)+1$

$=> 4cos^2(x)-1 = 0$

$=> cos^2(x)-1/4 = 0$

$=> (cos(x)+1/2)(cos(x)-1/2) = 0$

$=> cos(x)+1/2 = 0 or cos(x)-1/2 = 0$

$=> cos(x) = -1/2 or cos(x) = 1/2$

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

$cos(x) = -1/2 <=> x in {(2pi)/3, (4pi)/3}$
$cos(x) = 1/2 <=> x in {pi/3, (5pi)/3}$

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

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

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