SOLUTION: Find the solutions between [0,2pi] sin^2 θ + 3 sin θ − 28 = 0

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Question 851083: Find the solutions between [0,2pi]
sin^2 θ + 3 sin θ − 28 = 0
Answer by lwsshak3(11628) About Me  (Show Source):
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Find the solutions between [0,2pi]
sin^2 θ + 3 sin θ − 28 = 0
(sin θ+7)(sin θ-4)=0
sin θ=-7 (no solution, (-1≤sin θ≤1)
or
sin θ=4 (no solution, (-1≤sin θ≤1)