1. ## Find x

Find x :
$\left\{ \begin{array}{l}
\frac{1}{2}e^{x^2 - \frac{1}{2}} = \sin x \\
\frac{{\sqrt[4]{3}}}{{\sqrt 2 }}e^{\frac{{x^2 }}{2} - \frac{1}{4}} = \sqrt {\cos x} \\
\end{array} \right.
$

2. Use the identity $\sin^2x+\cos^2x=1$

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

$e^{2x^2-1}=1\Rightarrow 2x^2-1=0\Rightarrow x=\pm\frac{\sqrt{2}}{2}$

3. Hello, dhiab!

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

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

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?

4. i agree with soroban's observations, sort of two equation in one variable? unlikely occurence sort of . . . .

5. 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?

6. Originally Posted by eXist
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.