SOLUTION: how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.
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Question 609369: how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! how do you solve cos(x/2)-sinx=0 with the restriction of [0,2pi)? We are supposed to be using the half angle formula.
cos half-angle formula:
cos x/2=√[(1+cosx)/2]
√[(1+cosx)/2]-sinx=0
√[(1+cosx)/2]=sinx
square both sides
(1+cosx)/2=sin^2x=1-cos^2x
1+cosx=2-2cos^2x
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
..
2cosx-1=0
cosx=1/2
x=π/3 and 5π/3
or
cosx+1=0
cosx=-1
x=π
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