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joshuanoble2000

  • 4 years ago

solve the equation on the interval [0, 2pi) 8 sin^2x = -17 sinx - 9

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  1. y2o2
    • 4 years ago
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    8sin²(x)+17sin(x)+9 = 0 [8sin(x) + 9][sin(x) + 1] = 0 sin(x) = -9/8 or sin(x) = -1

  2. joshuanoble2000
    • 4 years ago
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    At this point do I take the arcsin of -9/8 and of -1 to find every possible x in the domain?

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