SOLUTION: Solve for theta 0 degrees < theta < 360 degrees. Sin(theta/2)+cos(theta)=1
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Question 770559: Solve for theta 0 degrees < theta < 360 degrees. Sin(theta/2)+cos(theta)=1
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ +\ \cos(\theta)\ =\ 1)
on the interval 
Half angle formula: \ =\ \pm\ \sqrt{\frac{1\ -\ \cos\left(\varphi\right) }{2}})
Substitute:
 }{2}}\ +\ \cos(\theta)\ =\ 1)
Isolate the radical
\ =\ \pm\ \sqrt{\frac{1\ -\ \cos\left(\varphi\right) }{2}})
Square both sides:
\ -\ 2\cos\left(\theta\right)\ =\ \frac{1\ -\ \cos\left(\theta\right)}{2})
Plain ol' algebra:
\ -\ 3\cos\left(\theta\right)\ +\ 1\ =\ 0)
Factor the quadratic:
\ -\ 1\right)\left(\cos\left(\theta\right)\ -\ 1\right)\ =\ 0)
Hence, \ =\ \frac{1}{2})
or \ =\ 1)
Check the unit circle: 
or
.
But )
, therefore:
or 
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve for theta 0 degrees < theta < 360 degrees. Sin(theta/2)+cos(theta)=1
***
sin(x/2)+cosx=1
sin(x/2)=±√((1-cosx)/2)
sin(x/2)=1-cosx
√((1-cosx)/2)=1-cosx
square both sides
(1-cosx)/2=1-2cosx+cos^2x
1-cosx=2-4cosx+2cos^2x
2cos^2x-3cosx+1=0
(2cosx-1)(cosx-1)=0
..
2cosx-1=0
cosx=1/2
x=π/3, 5π/3
..
cosx-1=0
cosx=1
x=0
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