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

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Question 630941: how do you solve sin(x/2)+cosx=0 with a restriction of [0,2pi)?
Found 2 solutions by lwsshak3, Barbazzo:
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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Identity: sin(x/2)=±√[(1-cosx)/2]
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
..
2cosx+1=0
cosx=-1/2
x=2π/3 and 4π/3 (in quadrants II and III where cos<0)
or
cosx-1=0
cosx=1
x=0


Answer by Barbazzo(2) About Me  (Show Source):
You can put this solution on YOUR website!
sin%28x%2F2%29%2Bcos+x+=+0
Transform Cos into Sin using Double Angle Identity: cos%282u%29+=+1-2sin%5E2%28u%29
Let u+=+x%2F2, so cos%28x%29+=+1-2sin%5E2%28x%2F2%29
sin%28x%2F2%29%2B1-2sin%5E2%28x%2F2%29+=+0
-2sin%5E2%28x%2F2%29%2Bsin%28x%2F2%29%2B1+=+0
2sin%5E2%28x%2F2%29-sin%28x%2F2%29-1+=+0
Factor: %28sin%28x%2F2%29-1%29%282sin%28x%2F2%29%2B1%29+=+0
Split into two equations:
sin%28x%2F2%29-1+=+0 and 2sin%28x%2F2%29%2B1+=+0
For sin%28x%2F2%29-1+=+0:
sin%28x%2F2%29+=+1
Take the inverse Sin:
x%2F2+=+pi%2F2
x+=+pi
For 2sin%28x%2F2%29%2B1+=+0:
2sin%28x%2F2%29+=+-1
sin%28x%2F2%29+=+-1%2F2
Take the inverse Sin:
x%2F2+=+7pi%2F6
x+=+7pi%2F3, but this is not within the restriction 0 to 2pi
OR
x%2F2+=+11pi%2F6
x+=+11pi%2F3, but this is not within the restriction 0 to 2pi
So the only value of x within the restriction 0 to 2pi is pi