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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=144 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 3, 0.6. Here's your graph:

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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
===========================
4*2.5 + 9*(5/3) = 10 + 15
= 25
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