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    A curve has parametric equations

    x= 3 - 2 sin t , y= 2t + 4 cos t

    I have shown that dy/dx = 2 tan t - sec t

    How do I now find all the values of t in the interval 0< t <2pi for which the curve has stationary points??

    I have put 2 tan t - sec t = 0 but I have no idea where to go from here. Can someone help me pleeeaaassseee?? :o:
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    Write out tan t and sec t in terms of sines and cosines.
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    2tant - sect = \frac {2sint}{cost} - \frac {1}{cost} = 0

    2sint - 1 = 0
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    (Original post by Pheylan)
    2tant - sect = \frac {2sint}{cost} - \frac {1}{cost} = 0

    2sint - 1 = 0
    Are you allowed to get rid of the cos t's like that?? you can multiply them by 0 to get 0?? If you can do that then brill and thank you because that has helped lots
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    (Original post by HaNzY)
    you can multiply them by 0 to get 0??
    well you can, but that would be a little pointless :awesome: i assume you meant multiply by cost? which you can indeed do
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    (Original post by Pheylan)
    well you can, but that would be a little pointless :awesome: i assume you meant multiply by cost? which you can indeed do
    Lol no i meant you can drag the cost to the other side and multiply it by the 0 thats over there and then they disappear? I think that's what you said but you said it more mathematically lol. Okay so I have the one value of pi/6, can you just remind me how I go about obtaining the other values between 0 and 2pi? I have really forgotten all this stuff! Thanks.
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    (Original post by HaNzY)
    Lol no i meant you can drag the cost to the other side and multiply it by the 0 thats over there and then they disappear? I think that's what you said but you said it more mathematically lol. Okay so I have the one value of pi/6, can you just remind me how I go about obtaining the other values between 0 and 2pi? I have really forgotten all this stuff! Thanks.
    sin(\pi - x) \equiv sinx
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    (Original post by Pheylan)
    sin(\pi - x) \equiv sinx
    So thats how they have their answer of 5pi/6 because they have done pi - 5pi/6. Ahhh its slowly coming back thank you, I remember we used to draw a mini graph of sin and then draw a line at 1/2 or whatever, to see how many solutions there were. However when I draw my graph now, I see no stationary points at pi/6 and 5pi/6. I guess thats because the curve is actually more complicated... Yeah lol I think I understand now, thank you :p:
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    (Original post by HaNzY)
    So thats how they have their answer of 5pi/6 because they have done pi - 5pi/6. Ahhh its slowly coming back thank you, I remember we used to draw a mini graph of sin and then draw a line at 1/2 or whatever, to see how many solutions there were. However when I draw my graph now, I see no stationary points at pi/6 and 5pi/6. I guess thats because the curve is actually more complicated... Yeah lol I think I understand now, thank you :p:
    i like to use the graph method too. you shouldn't look for stationary points at pi/6 or anything, look where sint = 1/2. then find the corresponding value of t
 
 
 
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