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Math Help - Trig equation

  1. #1
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    Trig equation

    apply sum-to-product identity to solve equation for [0,360)
    cos(t)-cos3(t)=0
    the book says that the answer is 0,90,180,270,360,and 540.
    I am getting 0,90,180,and 270, but thought that it was suppose to be less than 360.

    sin(2t)-sin(5t)=0 [0,360)

    i get -2cos(7t/2)sin(3t/2) and am totally lost on how to find the solutions.Do i divide 0 & 360 by 7/2 & 3/2 to find the solutions?The book has 0,pi/7,3pi/7,2pi/3,4pi/3, and 2pi.

    Can anyone help me and explain what i am doing wrong?
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  2. #2
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    Quote Originally Posted by Thelema View Post
    apply sum-to-product identity to solve equation for [0,360)
    cos(t)-cos3(t)=0
    the book says that the answer is 0,90,180,270,360,and 540.
    I am getting 0,90,180,and 270, but thought that it was suppose to be less than 360.
    I think you are right

    sin(2t)-sin(5t)=0 [0,360)

    i get -2cos(7t/2)sin(3t/2) and am totally lost on how to find the solutions.Do i divide 0 & 360 by 7/2 & 3/2 to find the solutions?The book has 0,pi/7,3pi/7,2pi/3,4pi/3, and 2pi.

    Can anyone help me and explain what i am doing wrong?
    Did you know that the general solution to sin(x) = sin(y) is x = n\pi + (-1)^n y?(n is an integer)
    [If you do not know the above think about it, its simple]


    Then all you have to do now is sin(2t) = sin(5t)

    So general solution is 2t = = n\pi + (-1)^n 5t

    n = 0 \Rightarrow 2t = 5t \Rightarrow t = 0

    n = 1 \Rightarrow 2t = \pi - 5t \Rightarrow t = \frac{\pi}7

    n = 2 \Rightarrow 2t = 2\pi + 5t \Rightarrow t = -\frac{2\pi}3 \,\, or \,\, t = \frac{4\pi}{3}

    n = 3 \Rightarrow 2t = 3\pi - 5t \Rightarrow t = \frac{3\pi}7

    Well I hope you get the idea
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  3. #3
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    Thank you very much. I have not learned your method yet, how would i do it
    0>7t/2>2pi?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Thelema View Post
    Thank you very much. I have not learned your method yet, how would i do it
    0>7t/2>2pi?
    First observe that, for any trig function if x is a solution, then 2pi+x,4pi+x,6pi+x,...... is a solution.

    Now observe that sin x = sin (pi - x). Which means if you get a solution with x as the answer, immediately (pi-x) qualifies too.

    We can always add or subtract 2pi (that is 360 degrees OR one full rotation) to any angle.So in summary, if x is a solution, then (pi-x), (2pi+x), (3pi-x), (4pi+x),..... are all solutions.Look at the pattern, and guess the general form.

    If x is a solution for a sin equation, then the generalised form solution is n\pi + (-1)^n x

    The best part about this is that you have to just obtain one solution in [0, \frac{\pi}2], Then you can plug in that solution as x in the above formula to get all possible solutions.
    That is exactly what I did in the previous post with 5t and 2t.
    Got it?
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