Finding angle in the isosceles triangle There is the following triangle. I infer that the triangle is isosceles. But I cannot go further.

 A: $AN = 5$ (Pythagoras), so $AC = 10 = AB$, which is how you can conclude $\triangle ABC$ is isosceles.
$\angle BAC = \angle MAN = \arctan\frac 34$, so $\angle CBA = \frac 12(180^{\circ} - \arctan \frac 34)$
$\angle MBN = \arctan \frac 36 = \arctan \frac 12$
$\therefore x = \frac 12(180^{\circ} - \arctan \frac 34) - \arctan \frac 12$
and you get a very nice answer in degrees.
EDIT:
The answer is $45^{\circ} = \frac{\pi}{4}$, where the latter is expressed in radian measure.
This answer used trigonometry. It is being left intact because the OP found it useful. However, there was a request from the OP in comments to see if a "by-hand" calculator free solution can be obtained with my approach. Indeed it can, and here is how:
The major problems to resolve are to try to reduce the arctangent terms to single inverse trigonometric ratios.
Rearranging the above,
$\displaystyle x = 90^{\circ} - (\frac 12\arctan\frac 34 + \arctan\frac 12)$
$\displaystyle \tan x = \tan (90^{\circ} - (\frac 12\arctan\frac 34 + \arctan\frac 12))= \frac 1{\tan(\frac 12\arctan\frac 34 + \arctan\frac 12)}$
which uses the identity $\tan (90^{\circ} - \theta) = \cot \theta = \frac 1{\tan\theta}$
Let's now focus on this expression: $\frac 12\arctan\frac 34$. This can be expressed as the arctangent of simple fraction as follows:
First let's find the tangent of the above. Note that $\tan \frac 12\theta = \frac{\sin\theta}{1+\cos\theta}$, which can be looked up in a reference, or may be proved fairly easily by considering the individual half-angle identities for sine and cosine and dividing them.
Also note that if $\theta = \arctan \frac34$ then $\sin\theta = \frac 35$ and $\cos\theta = \frac 45$, which is quite trivial to show by considering the $3-4-5$ special right triangle.
Putting those together:
$\displaystyle \tan \frac 12\arctan\frac 34 = \frac{\sin (\arctan\frac 34)}{1+ \cos (\arctan\frac 34)} = \frac{\frac 35}{1+\frac 45} = \frac 13$
Therefore $\frac 12\arctan\frac 34 = \arctan \frac 13$
So $\displaystyle \tan x =\frac 1{\tan(\frac 12\arctan\frac 34 + \arctan\frac 12)} = \frac 1{\tan(\arctan \frac 13 + \arctan \frac 12)}$
Apply the angle addition formula for tangent to get:
$\displaystyle \tan x = \frac 1{\frac{\tan(\arctan \frac 13) + \tan(\arctan \frac 12)}{1- \tan(\arctan \frac 13)\tan(\arctan \frac 12)}} = \frac 1{\frac{\frac 13+\frac 12}{1- \frac 13\cdot \frac 12}}  = \frac 11 = 1$
and the only acute angle solution for $\tan x = 1$ is $x = 45^{\circ} = \frac{\pi}{4}$.
I hope this is a more "satisfying" solution for the OP. As I said, it's possible to do it without a calculator, just harder.
A: Rotate $B$ by 180 degrees about $N$ to get point $B'$.  Drop perpendicular from $B'$ to $BC$ meeting $BC$ extended at $E$.

By Pythgoras, $NC=AN=5$ and $NB=3\sqrt5$, hence $AB=AC=10$, $BB'=6\sqrt{5}$.  Since $AB'CB$ is fixed under rotation by 180 degrees about $N$, it is a parallelogram.  So the parallelogram law gives $2AB^2+2BC^2=AC^2+BB'^2$, i.e., $BC=2\sqrt{10}$.
Since $AB=AC=CB'$, we have $BE=\frac{3}{2}BC=3\sqrt{10}$.  So again by Pythagoras, $B'E=\sqrt{BB'^2-BE^2}=\sqrt{90}=3\sqrt{10}=BE$.  So $\triangle BEB'$ is right-angled isosceles, hence $\angle NBC=\angle B'BE=45^\circ$.
A: Without trigonometry
Let $D$ be the symmetric of $C$ with respect to $A$, $H$ and $K$ the orthogonal projections of $A$ and $C$ on $BC$ and $AB$.
$BCD$ is a right triangle. By the midpoint theorem, $CK=2MN=6$ and $MN=MA=4$ so $BK=2$. By Pythagoras theorem, $BC=\sqrt{BK^2+CK^2}=2\sqrt{10}$ and $AH=\sqrt{AB^2-BH^2}=3\sqrt{10}$.
$\dfrac{NC}{ND}=\dfrac{5}{15}=\dfrac{1}{3}$ and $\dfrac{BC}{BD}=\dfrac{BC}{2AH}=\dfrac{1}{3}$ so, by the angle bisector theorem, $BN$ is the interior bisector of the right angle $CBD$. Thus $x=45^\circ$.

With trigonometry
$\cos BAC=4/5$ and $BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos BAC=40$ by the law of cosines.
$NC^2=NB^2+BC^2-2NB\cdot BC\cdot\cos x$ gives $x=45^{\circ}$
