Find $\lambda$ and $\theta$ such that it validates this matricial equation Find $\lambda$ and $\theta$  such that it validates the matricial equation
$$ \left( \begin{array}{cc}
1 & 2 \\
2 & 3
\end{array} \right)
%
\left( \begin{array}{cc}
\cos \theta \\
\sin \theta 
\end{array} \right) = \lambda\left( \begin{array}{cc}
\cos \theta \\
\sin \theta 
\end{array} \right)
$$

What I have tried:
$$\left\{ \begin{array}{ll}
         \cos \theta + 2\sin \theta  &= \lambda \cos \theta\\
        2\cos \theta + 3\sin \theta  &=  \lambda \sin \theta.\end{array} \right.  $$
$$\left\{ \begin{array}{ll}
         \ -2\cos \theta -4 \sin \theta  &= -2\lambda \cos \theta\\
        2\cos \theta + 3\sin \theta  &=  \lambda \sin \theta.\end{array} \right.  $$
$$-\sin \theta = -2\lambda \cos \theta + \lambda \sin \theta$$
$$ (\lambda + 1)\sin \theta = 2 \lambda \cos \theta$$
I can´t find $\lambda$ and $\theta$

For given $\theta$ this is an eigenvalue problem.
The characteristic equation of your matrix is $p(\lambda) = (1-\lambda)(3 - \lambda) -4 = \lambda^2 - 4\lambda -1$
And so the eigenvalues are 
$ \lambda  = \dfrac{4 \pm \sqrt{16+4}}{2} = 2 \pm \sqrt{5}$
It can be shown (though I used computer algebra) that the eigenvectors are
$$v_1 =  \begin{bmatrix} 1\\1 \end{bmatrix} v_2 =  \begin{bmatrix} \dfrac{2}{1+ \sqrt{5}} \\ \dfrac{2}{1-\sqrt{5}}\end{bmatrix}  $$
The polar coordinate representation of a vector allows you to find the angle.  Since there are two vectors, there should be two $\theta$, and so 
$$\theta_i = \tan ^{-1}(\dfrac{y_i}{x_i})$$
