SOLUTION: what is the least value of 4sec^2x+9cosec^2x ?

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Question 668858: what is the least value of 4sec^2x+9cosec^2x ?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
what is the least value of 4sec^2x+9cosec^2x ?
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f(x) = 4sec^2 + 9csc^ = 4cos^-2 + 9sin^-2
f'(x) = 4*-2cos^-3*(-sin) + 9*-2sin^-3*cos = 0
8sin/cos^3 -18cos/sin^3 = 0
8sin^4 - 18cos^4 = 0
4sin^4 - 9cos^4 = 0
cos^2 = 1-sin^2 --> cos^4 = sin^4 - 2sin^2 + 1
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4sin^4 - 9sin^4 + 18sin^2 - 9 = 0
5sin^4 - 18sin^2 + 9 = 0
Sub x for sin^2
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B-18x%2B9+=+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=%28-18%29%5E2-4%2A5%2A9=144.

Discriminant d=144 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--18%2B-sqrt%28+144+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-18%29%2Bsqrt%28+144+%29%29%2F2%5C5+=+3
x%5B2%5D+=+%28-%28-18%29-sqrt%28+144+%29%29%2F2%5C5+=+0.6

Quadratic expression 5x%5E2%2B-18x%2B9 can be factored:
5x%5E2%2B-18x%2B9+=+%28x-3%29%2A%28x-0.6%29
Again, the answer is: 3, 0.6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B-18%2Ax%2B9+%29

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Ignore the 3 solution, tho sin(x) = sqrt(3) has a complex solution.
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sin^2(x) = 0.6 --> cos^2(x) = 0.4
sec^2(x) = 2.5
csc^2(x) = 5/3
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4*2.5 + 9*(5/3) = 10 + 15
= 25