Two vectors having equal magnitude of x units acting at an angle of 45° have resultant $\sqrt{(2+\sqrt{2})}$ units. The value of x is ____ units.

Let two vector is $\overrightarrow{A}$ and $\overrightarrow{B}$
So, $\overrightarrow{A}$= x units, $\overrightarrow{B}$ = x units, $\theta =45°,\overrightarrow{R}=\sqrt{(2+\sqrt{2})}$
$\begin{array}{l}\overrightarrow{R}=\sqrt{{\left(\overrightarrow{A}\right)}^2+{\left(\overrightarrow{B}\right)}^2+2\overrightarrow{A}\overrightarrow{B}\cos \theta }\\ \sqrt{2+\sqrt{2}}=\sqrt{{\left(x\right)}^2+{\left(x\right)}^2+2x.x\cos 45°}\\ \sqrt{2+\sqrt{2}}=\sqrt{2x^2+2x^2.\frac{1}{\sqrt{2}}}\\ \sqrt{2+\sqrt{2}}=\sqrt{2x^2+\sqrt{2}{x}^2}\end{array}$      
Squaring on both side

$\begin{array}{l}2+\sqrt{2}=2x^2+\sqrt{2}{x}^2\\ (2+\sqrt{2})=(2+\sqrt{2})x^2\\ x^2=\frac{2+\sqrt{2}}{2+\sqrt{2}}\\ x^2=1\\ x=1\end{array}$