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Question
find the coordinates of the points on the curve y=2-cosx/sinx in the range 0<x<2pi where the gradient is 8/3
If the curve is y = 2 - cosx/sinx, then that is y = 2 - ctnx.
However, I believe it really to be y = (2-cosx)/sinx.
If this is so, it still resembles the ctnx curve in that it has undefined points at
0, pi, and 2pi. It is concave down on 0 to pi and concave up on pi to 2pi.
The extreme points would be at the solution to 1 - (2-cosx)ctn²x = 0.
That is the same as 1 = (2-cosx)ctn²x. Multiplying both sides by tan²x gives
tan²x = 2 - cosx since tanx*ctnx = 1.
This can then be converted to a polynomial equation and solved since the left side is
tan²x = sin²x/cos²x = (1-cos²x)/cos²x, which makes the equation be
(1-cos²x)/cos²x = 2 - cosx. Let t = cosx, multiply by t², put all variables on one side,
and then we have a cubic in t. The only valid solutions would be those where -1<=t<=1.
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Scotto
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