Solve system of equations with $\sin$ and $\cos$ Solve system of equations
$\begin{cases}
3x^2 + \sin 2y - \cos y - 3 = 0 \\
x^3 - 3x - \sin y - \cos 2y + 3 = 0
\end{cases}$
I tried to use substitution $x = \cos t$ or sth, but I get literally nothing

The equations are $x^2=A$ and $x^3-3x=B$ where
$$A=\frac{3+\cos y - \sin 2y}{3},\quad B=\sin y +\cos 2y - 3.$$
Now note that $x^3-3x=x(x^2-3),$ so one can substitute the value from $x^2=A$ here, obtaining $x(A-3)=B,$ that is, $x=B/(A-3).$ Then putting this $x$ back into $x^2=A$ gives the relation, involving only $A,B,$ that
$$B^2=A(A-3)^2.\tag{1}$$
Here $A,B$ are trig functions of $y$ admitting period $2\pi,$ and a root finder indicates there are four solutions for $y$ namely $y=0.066078,\ 0.523599,\ 2.61799,\ 3.06375.$
In degrees, the second and third appear to be 30 and 150 respectively, while the first and last seem about 3.785 and 175.54 degrees, not that "nice" as solutions. (I haven't checked analytically whether the 30 and 150 degree solutions are exactly correct.)
Anyway, for each of the four possible $y$ values, one can compute its corresponding $x$ from $x=B/(A-3)$ 
Added: I checked the $y$ being 30 and 150 degree solutions and they are exact, each leading to $x=1.$ That is, in radians, two of the four solutions are $(x,y)=(1,\pi/6),\ (1,5\pi/6).$ Also the two other solutions for $y$ appear to add up to $\pi$ and each of those leads to the same value of $x$ for the other two pairs $(x,y)$ of solutions, though these others are not particularly nice. Going with the equation $(1)$ and trying to say put all in terms of cosine gives a high degree polynomial to solve, I think degree 8 once one uses $\sin^2 y + \cos^2 y=1$ to eliminate the other trig function.
Edit: Correction-- the "other two $y$ solutions" (besides $\pi/6,5\pi/6$) in fact do not add to $\pi$ but to about $3.1298.$ Furthermore their corresponding $x$ values are about $1.1352$ from the smaller $y$ and $0.8481$ from the larger. I noticed this while trying (unsuccessfully) to show that $B^2-A(A-3)^2$ was symmetric around $\pi/2$; it wasn't.
