Question 746536: Solve the equation 4sin^2x-1=0 for x over the interval [0, 2π)
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! 4sin^2x - 1 + 0
4 sin^2x = 1
sin^x = 1/4
Take square root of both sides
sin x = + and - 1/2
Because the square root has been taken of the
angle both positive and negative values must
be included.
x = 30, 150, 210, 330, 390, 510,
570, 690, 750, 870, 930.
[ interval 0, 960}
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