SOLUTION: Find all the values of x in the interval [0,2pi] such that the gradient of the curve y=12sinx-5cosx is 13/2. (using differentiation)

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Question 1028991: Find all the values of x in the interval [0,2pi] such that the gradient of the curve y=12sinx-5cosx is 13/2. (using differentiation)
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the values of x in the interval [0,2pi] such that the gradient of the curve y=12sinx-5cosx is 13/2.
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y' = 12cos(x) + 5sin(x) = 13/2
12*sqrt(1 - sin^2) = -5sin + 13/2
144 - 144sin^2 = 25sin^2 - 65sin + 169/4
169sin^2 - 65sin - 407/4 = 0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 169x%5E2%2B-65x%2B-101.75+=+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=%28-65%29%5E2-4%2A169%2A-101.75=73008.

Discriminant d=73008 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--65%2B-sqrt%28+73008+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-65%29%2Bsqrt%28+73008+%29%29%2F2%5C169+=+0.991715757339482
x%5B2%5D+=+%28-%28-65%29-sqrt%28+73008+%29%29%2F2%5C169+=+-0.607100372724097

Quadratic expression 169x%5E2%2B-65x%2B-101.75 can be factored:
169x%5E2%2B-65x%2B-101.75+=+%28x-0.991715757339482%29%2A%28x--0.607100372724097%29
Again, the answer is: 0.991715757339482, -0.607100372724097. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+169%2Ax%5E2%2B-65%2Ax%2B-101.75+%29

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sin(x) = 0.991715757339482
x =~ 82.62 degs, 97.38 degs
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sin(x) = -0.607100372724097
x =~ 217.38 degs, 322.62 degs
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Convert those to radians.