Why this system is not linear and how that influences the number of solutions? I have an exercise in my last assignment for linear algebra, it's the following:

This system is not linear, in some sense
$$ 4sin(\alpha) + 3tan(\beta) + 2cos(\gamma) = 1 \\
 -sin(\alpha) + 2tan(\beta) - cos(\gamma) = 2 \\ sin(\alpha) + 2tan(\beta) - 3cos(\gamma) = -2 \\ $$
and yet we can nonetheless apply Gauss method. Do so. Does the
  system have no, one or many solutions? Explain.

What I have tried:
I tried to use a matrix representing the system. I replaced all $sin(\alpha)$ by $x$, all $tan(\beta)$ by $y$ and $ cos(\gamma)$ by $z$. This is the resulting matrix:
$$A = 
\left[\begin{array}{ccc|c}
4 & 3 & 2 & 1\\
-1 & 2 & -1 & 2 \\
1 & 2 & -3 & -2
\end{array}\right]
$$
I have tried to applied the Gauss method, multiplying the matrix $A$ by other matrices to make the matrix $A$ an upper triangular matrix. This is the resulting matrix, which I call $B$ (if you want to know all the steps, I could also type them all in LaTex/Markdown...):
$$B = 
\left[\begin{array}{ccc|c}
4 & 3 & 2 & 1\\
0 & 4 & -4 & 0 \\
0 & 0 & \frac{-9}{4} & \frac{1}{2}
\end{array}\right]
$$
Questions:
I don't know if this matrix is correct, but based on this matrix, how many solutions would I have: none, one or many? How can we say in general if a system has none, one or many solutions?
From my understanding, in my case, I would have one solution, because we have a "normal" upper triangular matrix. What about the other cases?
Can you also tell me if my resulting matrix is correct or not?
 A: The good news about solving systems of equations is that you can check them.
You have set up a system for $x,y,z$. You can check to see if your system is correct by solving for $x,y,z$, and then seeing if those solutions are solutions to the original system. Consider this empowering! You can check all of your work on this type of problem, all the time!
Something you must keep in mind, though, is that even though you might have a unique solution for $x$, your actual system is in $\alpha, \beta,$ and $\gamma$. Let us suppose that you found that $1 = x = \sin \alpha$ is the solution. Then $\alpha = \pi/2$ is a solution, but so is $\alpha = 2\pi + \pi/2$, and so on.
A: *

*If that matrix $B$ were correct, you would have only one solution. Since you started with a square matrix of coefficients ($3$ by $3$, not counting the ones to the right of the equals sign), The number of "free variables" in your solution equals the number of zero rows in the reduced, upper-triangular matrix before the augment. In your case that is zero, so there are no "free" variables. If you had any zero rows with a non-zero constant after the vertical augment, there would be no solutions; any zero rows with a zero constant would show infinitely many solutions. In a case like yours, no zero rows before the augment, there is exactly one solution.

*Your resulting matrix $B$ is not correct. Finding an upper-triangular matrix is not unique, so it is best to convert to an identity matrix, if possible, before the augment. This just takes more steps in the elimination process and results in the "reduced row echelon form." These are unique, so they are good for checking your upper triangular matrices. In your case your matrix $A$ reduces to
$$rref(A) = 
\left[\begin{array}{ccc|c}
1 & 0 & 0 & -1\\
0 & 1 & 0 &  1 \\
0 & 0 & 1 & 1
\end{array}\right]$$
But for $B$,
$$rref(B) = 
\left[\begin{array}{ccc|c}
1 & 0 & 0 & 19/36 \\
0 & 1 & 0 & -2/9 \\
0 & 0 & 1 & -2/9
\end{array}\right]$$
So you made a mistake somewhere. (I got the reduced row echelon forms from my graphing calculator but checked it manually.)
A note: Your initial matrix $A$ does not agree with your original equations. If your equations are correct, your $A$ should be
$$A=
\left[\begin{array}{ccc|c}
4 & 3 & 2 & 1\\
-1 & 2 & -1 & 2 \\
1 & 2 & 3 & -2
\end{array}\right]$$
I.e. the $3$ in the last row should be just $3$, not $-3$. Check which is correct: the equations or your matrix $A$.
Another note: your original equations are not linear in $\alpha, \beta,  \gamma$, but they are linear in $\sin\alpha, \tan\beta, \cos\gamma$. So your overall approach was correct. "Linear" in $\alpha, \beta,  \gamma$ here means that if $\alpha_1, \beta_1,  \gamma_1$ and $\alpha_2, \beta_2,  \gamma_2$ are solutions to your equations with the right hand sides changed to zeros, then another solution to those changed equations would be $r\alpha_1+s\alpha_2, r\beta_1+s\beta_2, r\gamma_1+s\gamma_2$ for any real numbers $r$ and $s$. (Some textbooks give two simpler conditions rather than this one complicated one, but the definitions are equivalent.) These kind of equations are called "linear" because if you graph one of these equations in the two-dimensional case the graph is a straight line. In your three-dimensional case each equation yields a flat plane, but it is still called linear. Your original equations do not satisfy that condition, but if you replace $\sin\alpha, \tan\beta, \cos\gamma$ with variables the new equations do satify it.
