SOLUTION: Approximate to the nearest 10' the solutions of the equation in the interval (o, 360). sin square (t)-8 sin(t)+3=0

Question 911843: Approximate to the nearest 10' the solutions of the equation in the interval (o, 360).
sin square (t)-8 sin(t)+3=0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Approximate to the nearest 10' the solutions of the equation in the interval (o, 360).
sin^2(t)-8 sin(t)+3 = 0
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Use the Quadratic Formula to get::
sin(t) = [8 +- sqrt(64-4*1*3)]/2 = [8+- sqrt(52)]/2
sin(t) = 7.606.. (extraneous)
sin(t) = 0.3944
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Ans::: t = arcsin(0.3944) = 23.23 degrees = 23 deg 14 minutes
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Cheers,
Stan H.
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