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Math Help - Find x

  1. #1
    Super Member dhiab's Avatar
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    Find x

    Find x :
    \left\{ \begin{array}{l}<br />
\frac{1}{2}e^{x^2 - \frac{1}{2}} = \sin x \\ <br />
\frac{{\sqrt[4]{3}}}{{\sqrt 2 }}e^{\frac{{x^2 }}{2} - \frac{1}{4}} = \sqrt {\cos x} \\ <br />
\end{array} \right.<br />
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  2. #2
    MHF Contributor red_dog's Avatar
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    Use the identity \sin^2x+\cos^2x=1

    Then \frac{1}{4}e^{2x^2-1}+\frac{3}{4}e^{2x^2-1}=1

    e^{2x^2-1}=1\Rightarrow 2x^2-1=0\Rightarrow x=\pm\frac{\sqrt{2}}{2}
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  3. #3
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    Hello, dhiab!

    I'm getting a totally different answer . . . What's going on?


    Find x\!:\begin{array}{cccc}<br />
\dfrac{1}{2}\,e^{\left(x^2 - \frac{1}{2}\right)} &=& \sin x & {\color{blue}[1]} \\ \\[-3mm]<br />
\dfrac{{\sqrt[4]{3}}}{{\sqrt 2 }}\,e^{\left(\frac{{x^2 }}{2} - \frac{1}{4}\right)} &=& \sqrt {\cos x} & {\color{blue}[2]}\\ \end{array}

    Square [2]: . \frac{\sqrt{3}}{2}\,e^{\left(x^2-\frac{1}{2}\right)} \:=\;\cos x\;\;[3]

    Divide [1] by [3]: . \frac{\sin x}{\cos x} \;=\;\frac{\frac{1}{2}e^{(x^2-\frac{1}{2})}} {\frac{\sqrt{3}}{2}e^{(x^2-\frac{1}{2})}} . \Rightarrow\quad \tan x \:=\:\frac{1}{\sqrt{3}}


    . . Therefore: . \boxed{x \;=\:\frac{\pi}{6}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I suspect the problem is in the given equations.

    Given a system of two equations, there should be two variables.

    Is there a typo, dhiab?

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  4. #4
    Senior Member pacman's Avatar
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    i agree with soroban's observations, sort of two equation in one variable? unlikely occurence sort of . . . .
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  5. #5
    Member eXist's Avatar
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    What if you were just given one of those equations, say:

    \frac{1}{2}e^{x^2 - \frac{1}{2}} = \sin x

    How then would one solve this?
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  6. #6
    MHF Contributor
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    Quote Originally Posted by eXist View Post
    What if you were just given one of those equations, say:

    \frac{1}{2}e^{x^2 - \frac{1}{2}} = \sin x

    How then would one solve this?
    with technology.
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