SOLUTION: Need help solving: sin^2(x/2)-2 = 0 and sec(x/2) = cos(x/2) for exact solution over the interval [0,2pi]. Thx!

Algebra ->  Trigonometry-basics -> SOLUTION: Need help solving: sin^2(x/2)-2 = 0 and sec(x/2) = cos(x/2) for exact solution over the interval [0,2pi]. Thx!       Log On


   

Question 861353: Need help solving:
sin^2(x/2)-2 = 0 and
sec(x/2) = cos(x/2)
for exact solution over the interval [0,2pi].
Thx!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
sin^2(x/2)-2 = 0
sin^2(x/2) = 2
Ans: no solution since sin^2 cannot exceed +1.
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sec(x/2) = cos(x/2)
1/cos(x/2) = cos(x/2)
cos^2(x/2) = 1
x/2 = 0 or x/2 = 2pi
x = 0 or x = 4pi
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Cheers,
Stan H.
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