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) (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) (Show Source): You can put this solution on YOUR website!
Transform Cos into Sin using Double Angle Identity:
Let , so
Factor:
Split into two equations:
and
For :
Take the inverse Sin:
For :
Take the inverse Sin:
, but this is not within the restriction 0 to
OR
, but this is not within the restriction 0 to
So the only value of x within the restriction 0 to is
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