SOLUTION: Solve the equation 4sin^2 &#952; - 2 = 0 on the interval 0<&#952;<2pi.

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Question 457590: Solve the equation 4sin^2 θ - 2 = 0 on the interval 0<θ<2pi.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equation 4sin^2 x - 2 = 0 on the interval 0< x <2pi.
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4sin^2(x) - 2 = 0
sin^2(x) = 1/2
sin(x) = +/-sqrt(2)/2
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In the interval 0 < x < 2pi, x = pi/4 or x = (3/4)pi
or x = (5/4)pi or (7/4)pi
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Cheers,
Stan H.
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