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joshuanoble2000
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solve the equation on the interval [0, 2pi)
8 sin^2x = 17 sinx  9
 2 years ago
 2 years ago
joshuanoble2000 Group Title
solve the equation on the interval [0, 2pi) 8 sin^2x = 17 sinx  9
 2 years ago
 2 years ago

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y2o2 Group TitleBest ResponseYou've already chosen the best response.1
8sin²(x)+17sin(x)+9 = 0 [8sin(x) + 9][sin(x) + 1] = 0 sin(x) = 9/8 or sin(x) = 1
 2 years ago

joshuanoble2000 Group TitleBest ResponseYou've already chosen the best response.0
At this point do I take the arcsin of 9/8 and of 1 to find every possible x in the domain?
 2 years ago
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