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.
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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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*[invoke solve_quadratic_equation 169,-65,-101.75]
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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.