Trigonometry
posted by Nikki .
Use the halfangle identities to find all solutions on the interval [0,2pi) for the equation
cos^2(x) = sin^2(x/2)

sin (x/2 ) = + sqrt([1cos x]/2) if x/2 in q 1 or 2 thus if 0 < x < 2 pi
sin^2 (x/2) = [1cos x]/2
so
cos^2 x = [1  cos x]/2
2 cos^2 x = 1  cos x
2 cos^2 x + cos x = 1
let z = cos x
2 z^2 + z 1 = 0
(2 z  1)(z +1) = 0
z = cos x = 1/2
or cos x = 1
x = 60 deg (pi/3) or 60 deg (5 pi/3)
or x = 180 deg (pi)
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