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}#