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8sin^2 x + 6sin x = 9 Solve for 0 ≤ x ≤ 2π Restricted (0, 2π)
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\[\frac{ 9 }{ 2 } and \frac{ 3 }{ 2 }\]
It's supposed to be in radians, uhh... I know I need the quadratic formula for it, and I'm supposed to get two values to set equal to zero, then find those values on a unit circle. A similar problem might give me sin x - 1 = 0, so I'd solve that for sin x; sin x = 1, and then find where sin is equal to 1 on the unit circle, which would be my answer.
sinx = 3/4
Do you mind showing me how you got that answer?
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