Question 630941
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=2&#960;/3 and 4&#960;/3 (in quadrants II and III where cos<0)
or
cosx-1=0
cosx=1
x=0