LinkBack URL
About LinkBacksFind 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 />](http://latex.codecogs.com/png.latex?\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 />
)
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}](http://latex.codecogs.com/png.latex?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]: .
Divide [1] by [3]: ..
. . Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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?
  |
Originally Posted by eXist 