SOLUTION: solve 12sin^2(x)+2sin(x)-2=0 where x is in radians between 0 and 2pi

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Question 815616: solve 12sin^2(x)+2sin(x)-2=0 where x is in radians between 0 and 2pi
Answer by Alan3354(69443) About Me  (Show Source):
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solve 12sin^2(x)+2sin(x)-2=0 where x is in radians between 0 and 2pi
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12sin^2(x)+2sin(x)-2=0
6sin^2(x) + sin(x) - 1 = 0
(3sin + 1)*(2sin - 1) = 0
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sin = -1/3
sin = 1/2