SOLUTION: how do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)?

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Question 609367: how do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
how do you solve sin(x/2)-cosx=0 with a restriction of [0,2pi)?
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sin(x/2)-cosx=0
sin(x/2)=cosx
√[(1-cosx)/2]=cosx
Square both sides
(1-cosx)/2=cos^2x
1-cosx=2cos^2x
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
cosx=1/2
x=π/3 and 5π/3
or
cosx=-1
x=π