Solving equation different solutions The vector $b=\pmatrix{x1\\x2}$ gets roated with a matrix about $30 \deg $ 
what results in the vector $ \overline b =\pmatrix{6\\8} $
Now my task is to find the original vector coordinates of $b$
I started like this:
The transformation-matrix is $T =\pmatrix{cos(30) & -sin(30) \\ sin(30) & cos(30)}$ and:
$$\pmatrix{cos(30) & -sin(30) \\ sin(30) & cos(30)} * \pmatrix{x1\\x2} = \pmatrix{6\\8}$$
What means:
$$x1 *cos(30)+ x2* -sin(30) = 6$$
$$x1*sin(30) + x2 * cos(30)=8$$
Next i dissolved the two equations for $x1 = 1$:
$$x2 = {6-cos(30)\over -sin(30)} $$
$$x2 = {8-sin(30\over cos(30)}$$
When i type this into my calculator i get for the first $x2 = -10.267...$ and for the second equation $x2 = 8.6602...$
My question is: Why do i get different solutions for $x2$? Did i make a mistake or is my calculator the problem? Thanks
 A: You get two different answers because $x_1 = 1$ is not the correct solution. Your two equations define two lines in the plane; they intersect at a single point. When you pick $x_1 = 1$, you get the $x_2$ values for the point that lie on each line with $x$-coordinate 1. 
The real problem is that when you "dissolved" the two equations, you did it wrong. The first should say 
$$
x_2 = \frac{6 - x_1 \cos 30}{-\sin 30}
$$
Now you should write out the second one in a similar form. That gives two different expressions for $x_2$, which you can set equal to each other, and solve for $x_1$, and you'll be on your way. 
Following your comment: 
You cannot "do it for the first equation": you need to solve the two equations together. So far we have
$$
x_2 = \frac{6 - x_1 \cos 30}{-\sin 30}
$$
For the second -- 
$$
x_1 \sin 30 + x_2 \cos 30 = 8
$$
-- you can do the following steps:
\begin{align}
x_1 \sin 30 + x_2 \cos 30 &= 8\\
x_2 \cos 30 &= 8 - x_1 \sin 30 \\
x_2  &= \frac{8 - x_1 \sin 30}{\cos 30}.
\end{align}
Now we have two expressions for $x_2$ which we can set equal:
\begin{align}
x_2  &= \frac{8 - x_1 \sin 30}{\cos 30} \text{, and}\\
x_2 &= \frac{6 - x_1 \cos 30}{-\sin 30}\text{, so}\\
\frac{8 - x_1 \sin 30}{\cos 30} &= \frac{6 - x_1 \cos 30}{-\sin 30}.\\
(8 - x_1 \sin 30)(-\sin 30)&= (6 - x_1 \cos 30)({\cos 30} )\\
-8\sin 30  + x_1 \sin^2 30&= 6\cos 30 - x_1 \cos^2 30\\
x_1 \sin^2 30 + x_1 \cos^2 30&= 6\cos 30 + 8 \sin 30\\
x_1 (\sin^2 30 + \cos^2 30)&= 6\cos 30 + 8 \sin 30\\
x_1 &= 6\cos 30 + 8 \sin 30\\
~\approx 9.2.
\end{align}
Similarly, you can find that 
$$
x_2 = -6 \sin 30 + 8 \cos 30 \approx 3.93
$$
Note that these final computations of $x_1$ and $x_2$ are exactly those suggested in @user84413's answer; I worked out these details because I thought you might not be that familiar with the matrix mathematics yet. That other answer is clearly the way to go, once you get more familiar with stuff like this. 
A: Another way you could approach this problem, which might be a little easier, is to use the fact that the inverse matrix will rotate $\pmatrix{6\\8}$ back to $\pmatrix{x_{1}\\x_{2}}$.
Since the inverse matrix corresponds to a rotation of $-30^{\circ}$, you have that
$\pmatrix{x_{1}\\x_{2}}=\begin{pmatrix} \cos(-30^{\circ}) & -\sin(-30^{\circ})\\ \sin(-30^{\circ}) & \cos(-30^{\circ})\end{pmatrix}\pmatrix{6\\8}$ 
where $\cos(-30^{\circ})=\cos(30^{\circ})$ and $\sin(-30^{\circ})=-\sin(30^{\circ})$.
