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)

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) (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 (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=73008 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.991715757339482, -0.607100372724097. Here's your graph:

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

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