Results 1 to 13 of 13

Math Help - Solve for x

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Solve for x

    Solve the equation:
    \arcsin(\frac{x}{x-1}) +2 \arctan(\frac{1}{x+1})= \frac{\pi}{2}

    My attempt:
    let \theta = \arcsin(\frac{x}{x-1})......1
    and \phi =\arctan(\frac{1}{x+1}).......2
    from these:
    \sin\theta=\frac{x}{x-1}
    \tan\phi=\frac{1}{x+1}
    substitute 1 and 2 into the equation:
    \theta + 2 \phi = \frac{\pi}{2}
    \sin (\theta+2\phi) =\sin \frac{\pi}{2}
    \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1
    substitute \sin\theta and use the identity:
    \sin2\phi=\frac{2t}{1+t^2} and \cos2\phi=\frac{1-t^2}{1+t^2} where t=\tan\phi
    (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1
    (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x  +1})^2})=1
    my problem is the \cos\theta, I tried using pythagoras' theorem to find the the length of the adjacent side the then use that to determine the cosine, but I can't work it out.
    thanks for any help.
    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

    Trig equation

    Hello arze
    Quote Originally Posted by arze View Post
    Solve the equation:
    \arcsin(\frac{x}{x-1}) +2 \arctan(\frac{1}{x+1})= \frac{\pi}{2}

    My attempt:
    let \theta = \arcsin(\frac{x}{x-1})......1
    and \phi =\arctan(\frac{1}{x+1}).......2
    from these:
    \sin\theta=\frac{x}{x-1}
    \tan\phi=\frac{1}{x+1}
    substitute 1 and 2 into the equation:
    \theta + 2 \phi = \frac{\pi}{2} OK up to here - see below.
    \sin (\theta+2\phi) =\sin \frac{\pi}{2}
    \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1
    substitute \sin\theta and use the identity:
    \sin2\phi=\frac{2t}{1+t^2} and \cos2\phi=\frac{1-t^2}{1+t^2} where t=\tan\phi
    (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1
    (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x  +1})^2})=1
    my problem is the \cos\theta, I tried using pythagoras' theorem to find the the length of the adjacent side the then use that to determine the cosine, but I can't work it out.
    thanks for any help.
    From the point that I indicated:

    \theta = \tfrac{\pi}{2} - 2\phi

    \Rightarrow \sin\theta = \sin(\tfrac{\pi}{2} - 2\phi) = \cos2\phi

    \Rightarrow \frac{x}{x-1}=\frac{1-t^2}{1+t^2}

    \Rightarrow \frac{x}{x-1}=\frac{1-\frac{1}{(x+1)^2}}{1+\frac{1}{(x+1)^2}}

    = \frac{(x^2+2x+1)-1}{(x^2+2x+1)+1}

    =\frac{x(x+2)}{x^2+2x+2}

    \Rightarrow x = 0 or \frac{1}{x-1}=\frac{x+2}{x^2+2x+2}

    \Rightarrow x^2+2x+2 = (x-1)(x+2)

    \Rightarrow x = -4

    So the solutions are x = 0, -4

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    How did you get x=0? i don't understand how you arrived at that solution.
    also the answer is supposed to be only 0, but i think i know why.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    hi
    the equation x^2+2x+2 has no real roots.
    to solve \frac{x(x+2)}{x^2+2x+2} only x(x+2) must be equal to zero,so the solutions are \mathbb{S}=\left \{ 0,-2 \right \}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, arze!

    There must be a typo . . .


    Solve the equation:

    . . \arcsin\underbrace{\left(\frac{x}{x-1}\right)} +2 \arctan\left(\frac{1}{x+1}\right) \;=\; \frac{\pi}{2}
    . . . .  {\color{blue}^{\text{This is greater than 1}}}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    i think x must be a negative integer or zero
    In addition to "Soroban"
    Arcsin : [-1,1] \mapsto \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello arze
    Quote Originally Posted by arze View Post
    How did you get x=0? i don't understand how you arrived at that solution.
    also the answer is supposed to be only 0, but i think i know why.
    Because x is a factor of both sides of the equation \frac{x}{x-1}=\frac{x(x+2)}{x^2+2x+2}

    I'm not sure what the problem is with x = -4. I'll think about it!

    Grandad
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    i see! thanks! now the only problem i have left is the -4 value
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Quote Originally Posted by Soroban View Post
    Hello, arze!

    There must be a typo . . .
    i copied it exactly as it appears in the book, so i had concluded that x is 0 or negative
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    the only solution for x is x = 0; the value x = -4 is not possible.
    Last edited by pacman; July 28th 2009 at 12:14 AM. Reason: not visible
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello arze
    Quote Originally Posted by arze View Post
    i see! thanks! now the only problem i have left is the -4 value
    The problem of extra 'solutions' is always likely to occur when we square things as part of solving an equation, because squaring a real number always 'hides' a possible minus sign by giving a positive value, whatever the sign of the original number.

    It's pretty obvious when you do something like this, for instance:

    2x-3 = x+4

    \Rightarrow (2x-3)^2=(x+4)^2

    \Rightarrow 4x^2 -12x+9 =x^2+8x+16

    \Rightarrow 3x^2-20x-7=0

    \Rightarrow (3x+1)(x-7)=0

    \Rightarrow x=-\tfrac13 or x=7

    All of which is correct, but that doesn't mean that you can reverse the implication arrows and say that both values of x are valid solutions to the original equation.

    x=7 is OK. (This gives LHS = 11 = RHS.)

    But x = -\tfrac13 isn't. This gives LHS = -\tfrac{11}{3}, and RHS = +\tfrac{11}{3}.

    So, whereas x =-4 is a valid solution to the equation

    \frac{x}{x-1}=\frac{x(x+2)}{x^2+2x+2}

    it doesn't necessarily mean that it's a solution to the original equation. x=-4 gives \sin\theta = \tfrac45 and \tan\phi = -\tfrac13, whereas you'll find you need \sin\theta = \tfrac45,\,\tan\phi = +\tfrac13 to satisfy the original equation.

    The problem arises when we use \cos2\phi=\frac{1-t^2}{1+t^2}, because this has the same value, \tfrac45, whether t = \tfrac13 or t=-\tfrac13.

    So, we all need to take care with signs when we take squares and square roots!

    Grandad
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    You hammered the nail so well Grandad! That's the reason why x = -4 is not a solution to the problem, but only x = 0.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    you can see here, only x = 0.
    Attached Thumbnails Attached Thumbnails Solve for x-asso.gif  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. how do i solve this IVP: y'=y^2 -4
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2010, 11:14 AM
  3. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 02:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum