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Math Help - Help with solving a trigonometric equation

  1. #1
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    Help with solving a trigonometric equation

    I am trying to find the x-intercepts of this function.

     y=2\sin(3x) - 2 \in[-2\pi,2\pi]

    Okay, so I make the domain  3x \in[-6\pi,6\pi]

    x ints, y=0

    0=2\sin(3x) -2
    1=\sin(3x)

    Basic angle  = \sin^{-1}(1) = \frac{\pi}{2}

    \sin is positive in 1st + 2nd quadrants.

    3x=\sin^{-1}(1)
    3x=\frac{\pi}{2} (repeated) , \frac{5\pi}{2} (repeated) , \frac{9\pi}{2} (repeated)

    Dividing through by 3

    x=\frac{\pi}{6} (repeated) , \frac{5\pi}{6} (repeated) , \frac{3\pi}{2} (repeated)

    So these are the intercepts from [0,2\pi]

    My question is, how do I get the intercepts from [-2\pi,0] ?
    Last edited by user_5; October 14th 2009 at 10:05 PM. Reason: Fixed wording, edited question. My bad =/
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  2. #2
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    Hello user_5
    Quote Originally Posted by user_5 View Post
    Alright, so I am trying to solve the function for x:

     y=2\sin(3x) - 2 x \in[-2\pi,2\pi]
    I don't know what you mean by 'solve'. This is y written as a function of x.

    Okay, so I make the domain  3x \in[-6\pi,6\pi]

    x ints, y=0

    0=2\sin(3x) -2
    1=\sin(3x)
    I don't understand this either. You have put y = 0, but what about the term in 2x? This just appears as 2 now.

    So I don't think any of this:
    Basic angle  = \sin^{-1}(1) = \frac{\pi}{2}

    \sin is positive in 1st + 2nd quadrants.

    3x=\sin^{-1}(1)
    3x=\frac{\pi}{2} (repeated) , \frac{5\pi}{2} (repeated) , \frac{9\pi}{2} (repeated)

    Dividing through by 3

    x=\frac{\pi}{6} (repeated) , \frac{5\pi}{6} (repeated) , \frac{3\pi}{2} (repeated)

    So these are the intercepts from [0,2\pi]

    My question is, how do I get the intercepts from [-2\pi,0] ?
    makes much sense.

    Could we have the original question again please?

    Grandad

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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello user_5I don't know what you mean by 'solve'. This is y written as a function of x.
    Well, he said that he wants to solve the function for x. So I think he is trying to find the inverse of y. But I don't understand what he wrote after that either.

    Actually, I think that would be impossible. You can't isolate x.
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  4. #4
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    Hello adkinsjr
    Quote Originally Posted by adkinsjr View Post
    Well, he said that he wants to solve the function for x. So I think he is trying to find the inverse of y. But I don't understand what he wrote after that either.

    Actually, I think that would be impossible. You can't isolate x.
    Agreed. You cannot make x the subject of y = 2\sin3x -2x.

    That's why I don't understand what 'solve' means in this context.

    Grandad
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  5. #5
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    Sorry, I fixed the typo it is just -2 instead of -2x.
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  6. #6
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    Hello user_5
    Quote Originally Posted by user_5 View Post
    Sorry, I fixed the typo it is just -2 instead of -2x.
    OK, so you want to know the values of x between -2\pi and 0 that satisfy \sin3x = 1.

    You have already noted that if -2\pi \le x \le 0, then -6\pi \le3x \le0, so the values of 3x are:

    3x=-\frac{3\pi}{2},-\frac{7\pi}{2}, -\frac{11\pi}{2}

    \Rightarrow x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}

    Grandad
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  7. #7
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    Thanks, but how did you get those answers?
    <br />
x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}<br />

    I'm not exactly sure how to go in the clockwise direction of the unit circle.

    Like if you were going anti clockwise, the answer in the
    1st quadrant = Just the x value
    2nd quadrant = \pi - x
    3rd quadrant = \pi+ x
    4th quadrant = 2\pi- x
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  8. #8
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    Hello user_5
    Quote Originally Posted by user_5 View Post
    Thanks, but how did you get those answers?
    <br />
x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}<br />

    I'm not exactly sure how to go in the clockwise direction of the unit circle.

    Like if you were going anti clockwise, the answer in the
    1st quadrant = Just the x value
    2nd quadrant = \pi - x
    3rd quadrant = \pi+ x
    4th quadrant = 2\pi- x
    Using degrees rather than radians (it's easier!) start at 0^o at the point (1,0) on the unit circle. Rotate 90^o clockwise to the point (0,-1): y-value = -1. So \sin(-90^o) = -1.

    Rotate a further 90^o to -180^o: \sin(-180^o)=0

    A further 90^o to -270^o: \sin(-270^o) = 1. That's the first one we want!

    Every 360^o further takes you to the same point: (0,1).

    In radians, then, all the following have a sine of 1: -\frac{3\pi}{2},\,-\frac{3\pi}{2}-2\pi=-\frac{7\pi}{2},\,-\frac{7\pi}{2}-2\pi=-\frac{11\pi}{2}, ...


    Grandad
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