SOLUTION: Please help!! Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°). sin^2(t)-8sin(t)+1=0

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Question 993625: Please help!! Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0
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Sub x for sin(t)
x%5E2+-+8x+%2B+1+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-8%29%5E2-4%2A1%2A1=60.

Discriminant d=60 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+60+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+60+%29%29%2F2%5C1+=+7.87298334620742
x%5B2%5D+=+%28-%28-8%29-sqrt%28+60+%29%29%2F2%5C1+=+0.127016653792583

Quadratic expression 1x%5E2%2B-8x%2B1 can be factored:
1x%5E2%2B-8x%2B1+=+%28x-7.87298334620742%29%2A%28x-0.127016653792583%29
Again, the answer is: 7.87298334620742, 0.127016653792583. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B1+%29

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sin(t) =~ 0.12701665
t =~ 7º 20'
t =~ 172º 40'

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)-8sin(t)+1=0
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Use quadratic formula:
sin(t) = [8+-sqrt(64-4)]/2
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sin(t) = [8+-sqrt(60)]/2
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sin(t) = 8+-sqrt(15)
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Comment:: Both values for sin(t) are out of the range of the sin function.
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Ans: No solution.
Cheers,
Stan H.
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