Find stationary point of y^3 + 3y + 3 cos 2x + 3 sin x = 0

[Monkeyseat]

Hi,

Question

A curve has equation y^3 + 3y + 3 cos 2x + 3 sin x = 0.

a) Show that dy/dx = (2 sin 2x - cos x)/(y^2 + 1).
b) Hence find the coordinates of all the stationary points of the curve in the interval 0 < x < 2pi, giving all your answers to three significant figures.

Working

a)

I was ok with this part, so I won't bother posting working.

b)

I'm not sure what to do here however.

Ok, so the stationary point will be when dy/dx = 0.

dy/dx = (2 sin 2x - cos x)/(y^2 + 1)
0 = (2 sin 2x - cos x)/(y^2 + 1)
2 sin 2x - cos x = 0

But I don't know how I should solve that. I think I need to write it all in terms of either sin x or cos x. I thought that I may need to write it in the form R sin (2x - alpha), but I couldn't get this to work.

Instead I went on to do this:

2 sin 2x = cos x
4 sin^2 (2x) = cos^2 (x)
4 sin^2 (2x) = 1 - sin^2 (x)
5 sin^2 (2x) = 1
sin^2 (2x) = 1/5
sin x = +sqrt.(1/5) and -sqrt.(1/5)

However, when I use that it does not give me the correct answers.

Any help is greatly appreciated.

Thanks.

 

[skeeter]

\(\displaystyle 2\sin(2x) - cos{x} = 0\)

\(\displaystyle 2(2\sin{x}\cos{x}) - \cos{x} = 0\)

\(\displaystyle \cos{x}(4\sin{x} - 1) = 0\)

finish.

 

[Monkeyseat]

I see how to do it now. Thanks, I managed to finish the question from where you left it and get the right answers.

Thanks again.