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    I have got so much maths homework to do =(.. and i have just come across this question i can't do:

    y=sin^3x have five stationary points between -pie and pie.
    Find the five stationary points.

    I've worked out dy/dx to be: cosx x 3(sin x)^2
    and now im stuck

    Can anyone help?

    xxx
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    You know that the stationary points are when \frac{dy}{dx} = 0; so you need to solve 3cosxsin^2x = 0. There should be 5 solutions for x in between -\pi and \pi. Then plug those values back into the original equation to find y.
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    yeah i know that.. but i just dont know what the solutions would be:| x
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    solve cos x = 0 and sin^2 x = 0
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    Ah. Treat it like you would a factorised quadratic and solve it separately:
    cosx=0 and sin^2 x=0
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    i mite sound like im being stupid but how would i solve cosx=0 ? x
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    cosx = 0
    therefore one solution is x=cos^{-1}0

    If you put cos^{-1}0 into your calculator, you should get \frac{\pi}{2}. That's one of the answers.
    Do you know what the graph of y=cosx looks like? Sketch it for x between -\pi and \pi and use the symmetries of the graph to work out the other solutions. (You should get one other solution.)

    Do the same for \sin^2 x = 0, but square root it first to make it simpler. There should be three solutions for this one.
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    square root what?
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    Both sides of the equation, so it becomes sinx = 0.
    It'll work with sin^2 x=0, but most people are more familiar with sinx=0.
 
 
 

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