Question 1058897: Solve cos(x/2)=-sinx on the interval 0 to 2pi
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Solve cos(x/2)=-sinx on the interval 0 to 2pi
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cos(x/2) = -2sin(x/2)*cos(x/2)
cos(x/2) + 2sin(x/2)*cos(x/2) = 0
cos(x/2)*(1 + 2sin(x/2)) = 0
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cos(x/2) = 0
x/2 = pi/2, 3pi/2
--> x = pi
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1 + 2sin(x/2) = 0
2sin(x/2) = -1
sin(x/2) = -1/2
x/2 = 7pi/6, 11pi/6
--> x > 2pi
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