Results 1 to 8 of 8

Thread: Help with solving a trigonometric equation

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    38

    Help with solving a trigonometric equation

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

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

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

    x ints, y=0

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

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

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

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

    Dividing through by 3

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

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

    My question is, how do I get the intercepts from $\displaystyle [-2\pi,0]$ ?
    Last edited by user_5; Oct 14th 2009 at 09:05 PM. Reason: Fixed wording, edited question. My bad =/
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello user_5
    Quote Originally Posted by user_5 View Post
    Alright, so I am trying to solve the function for x:

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

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

    x ints, y=0

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

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

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

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

    Dividing through by 3

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

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

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

    Could we have the original question again please?

    Grandad

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    680
    Thanks
    19
    Quote Originally Posted by Grandad View Post
    Hello user_5I don't know what you mean by 'solve'. This is $\displaystyle y$ written as a function of $\displaystyle x$.
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    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 $\displaystyle x$ the subject of $\displaystyle y = 2\sin3x -2x$.

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2009
    Posts
    38
    Sorry, I fixed the typo it is just -2 instead of -2x.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    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 $\displaystyle x$ between $\displaystyle -2\pi$ and $\displaystyle 0$ that satisfy $\displaystyle \sin3x = 1$.

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

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

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2009
    Posts
    38
    Thanks, but how did you get those answers?
    $\displaystyle
    x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}
    $

    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 $\displaystyle x$ value
    2nd quadrant = $\displaystyle \pi - x$
    3rd quadrant = $\displaystyle \pi+ x$
    4th quadrant = $\displaystyle 2\pi- x$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello user_5
    Quote Originally Posted by user_5 View Post
    Thanks, but how did you get those answers?
    $\displaystyle
    x=-\frac{\pi}{2},-\frac{7\pi}{6},-\frac{11\pi}{6}
    $

    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 $\displaystyle x$ value
    2nd quadrant = $\displaystyle \pi - x$
    3rd quadrant = $\displaystyle \pi+ x$
    4th quadrant = $\displaystyle 2\pi- x$
    Using degrees rather than radians (it's easier!) start at $\displaystyle 0^o$ at the point $\displaystyle (1,0)$ on the unit circle. Rotate $\displaystyle 90^o$ clockwise to the point $\displaystyle (0,-1)$: $\displaystyle y$-value $\displaystyle = -1$. So $\displaystyle \sin(-90^o) = -1$.

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

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

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

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


    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Trigonometric Equation
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: May 17th 2011, 01:20 AM
  2. Help in Solving a Trigonometric Equation
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Aug 10th 2010, 12:01 AM
  3. Solving a trigonometric equation.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 26th 2010, 07:14 AM
  4. Solving Trigonometric equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 8th 2010, 10:12 PM
  5. Solving a trigonometric equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Mar 6th 2009, 03:54 PM

Search Tags


/mathhelpforum @mathhelpforum