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Thread: Solve for x

  1. #1
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    Solve for x

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

    My attempt:
    let $\displaystyle \theta = \arcsin(\frac{x}{x-1})$......1
    and $\displaystyle \phi =\arctan(\frac{1}{x+1})$.......2
    from these:
    $\displaystyle \sin\theta=\frac{x}{x-1}$
    $\displaystyle \tan\phi=\frac{1}{x+1}$
    substitute 1 and 2 into the equation:
    $\displaystyle \theta + 2 \phi = \frac{\pi}{2}$
    $\displaystyle \sin (\theta+2\phi) =\sin \frac{\pi}{2}$
    $\displaystyle \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1$
    substitute $\displaystyle \sin\theta$ and use the identity:
    $\displaystyle \sin2\phi=\frac{2t}{1+t^2}$ and $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$ where $\displaystyle t=\tan\phi$
    $\displaystyle (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1$
    $\displaystyle (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x +1})^2})=1$
    my problem is the $\displaystyle \cos\theta$, 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.
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  2. #2
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    Trig equation

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

    My attempt:
    let $\displaystyle \theta = \arcsin(\frac{x}{x-1})$......1
    and $\displaystyle \phi =\arctan(\frac{1}{x+1})$.......2
    from these:
    $\displaystyle \sin\theta=\frac{x}{x-1}$
    $\displaystyle \tan\phi=\frac{1}{x+1}$
    substitute 1 and 2 into the equation:
    $\displaystyle \theta + 2 \phi = \frac{\pi}{2}$ OK up to here - see below.
    $\displaystyle \sin (\theta+2\phi) =\sin \frac{\pi}{2}$
    $\displaystyle \sin\theta\cos2\phi + \cos\theta\sin2\phi = 1$
    substitute $\displaystyle \sin\theta$ and use the identity:
    $\displaystyle \sin2\phi=\frac{2t}{1+t^2}$ and $\displaystyle \cos2\phi=\frac{1-t^2}{1+t^2}$ where $\displaystyle t=\tan\phi$
    $\displaystyle (\sin\theta)(\frac{1-t^2}{1+t^2}) + (\sin\theta)(\frac{2t}{1+t^2})=1$
    $\displaystyle (\frac{x}{x-1})(\frac{1-(\frac{1}{x+1})^2}{1+(\frac{1}{x+1})^2}) + (\cos\theta)(\frac{2(\frac{1}{x+1})}{1+(\frac{1}{x +1})^2})=1$
    my problem is the $\displaystyle \cos\theta$, 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:

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

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

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

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

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

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

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

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

    $\displaystyle \Rightarrow x = -4$

    So the solutions are $\displaystyle x = 0, -4$

    Grandad
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  3. #3
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    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.
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  4. #4
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    hi
    the equation $\displaystyle x^2+2x+2 $ has no real roots.
    to solve $\displaystyle \frac{x(x+2)}{x^2+2x+2}$ only $\displaystyle x(x+2)$ must be equal to zero,so the solutions are $\displaystyle \mathbb{S}=\left \{ 0,-2 \right \}$
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  5. #5
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    Hello, arze!

    There must be a typo . . .


    Solve the equation:

    . . $\displaystyle \arcsin\underbrace{\left(\frac{x}{x-1}\right)} +2 \arctan\left(\frac{1}{x+1}\right) \;=\; \frac{\pi}{2}$
    . . . . $\displaystyle {\color{blue}^{\text{This is greater than 1}}} $
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  6. #6
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    i think $\displaystyle x$ must be a negative integer or zero
    In addition to "Soroban"
    $\displaystyle Arcsin : [-1,1] \mapsto \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$
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  7. #7
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    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 $\displaystyle x$ is a factor of both sides of the equation $\displaystyle \frac{x}{x-1}=\frac{x(x+2)}{x^2+2x+2}$

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

    Grandad
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  8. #8
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    i see! thanks! now the only problem i have left is the -4 value
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  9. #9
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    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
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  10. #10
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    the only solution for x is x = 0; the value x = -4 is not possible.
    Last edited by pacman; Jul 28th 2009 at 12:14 AM. Reason: not visible
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  11. #11
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    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:

    $\displaystyle 2x-3 = x+4$

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

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

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

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

    $\displaystyle \Rightarrow x=-\tfrac13$ or $\displaystyle x=7$

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

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

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

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

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

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

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

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

    Grandad
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  12. #12
    Senior Member pacman's Avatar
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    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.
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  13. #13
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    you can see here, only x = 0.
    Attached Thumbnails Attached Thumbnails Solve for x-asso.gif  
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