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

Invoking the double-angle formulas $\sin 2 y = 2 \sin y \cos y$ and $\cos 2 y = 1 - 2 \sin^2 y$, and abbreviating "$\sin y$" and "$\cos y$" by "$s$" and "$c$", we can manipulate the equations into these forms
$$\begin{align}
c \;( 2 s - 1 ) &= - 3 ( x^2 - 1 ) &= (x-1)\;P \tag{1}\\
s \;( 2 s - 1 ) &= -(x - 1 )^2 ( x + 2 ) &= (x-1)\;Q \tag{2}
\end{align}$$
where $P := -3(x+1)$ and $Q := -(x-1)(x+2)$. From here, we see a family of solutions that make each side of $(1)$ and $(2)$ vanish simultaneously: 

$x=1$, and all $y$ such that $\sin y = \frac12$ (in particular, $y = \frac\pi6$ and $y = \frac{5\pi}{6}$)

We also see that there are no other solutions with either $x=1$ or $\sin y = \frac12$. Moreover, there are no other solutions with either $P=0$ or $Q=0$: if $x=-1$, then $(2)$ gives an impossible relation, $c(2s-1)=-4$; likewise, if $x=-2$, then $(1)$ gives $c(2s-1) = -9$. Consequently, we can write
$$\frac{c}{P} = \frac{x-1}{2s-1} = \frac{s}{Q}$$
which implies
$$c = \pm\frac{P}{R}\qquad s = \pm\frac{Q}{R}$$
where $R := \sqrt{P^2+Q^2}$ and the "$\pm$"s match.
Substituting into $(1)$ (or $(2)$), then,
$$2 Q \mp R = ( x - 1 ) R^2 \quad\to\quad \left(\;2 Q - ( x - 1 )R^2\;\right)^2 = R^2$$
Rewriting $Q$ and $R^2 = P^2 + Q^2$ in terms of $x$, we have
$$x^{10}+ 2 x^9 + 9 x^8 + 28 x^7 + 38 x^6 + 48 x^5 + 73 x^4 - 118 x^3 - 345 x^2 - 48 x + 276 = 0$$
for which Mathematica finds two real roots

$$x_1 \approx 1.1352 \qquad x_2 \approx 0.848153$$

The corresponding values of $Q/P$ give the tangents of the associated $y$s:
$$\tan y_1 = \frac{-0.423884}{-6.4056} = \phantom{-}0.0661739$$
$$\tan y_2 = \frac{\phantom{-}0.432483}{-5.54446} = -0.0780028$$
One can check that 

Only first-quadrant solutions $y_1$ and second-quadrant solutions $y_2$ actually satisfy the original equations.

Here's a graph of one period of the situation:

Equation $(1)$ defines the blue squiggles (right and middle); equation $(2)$ defines the red squiggle (left) and ovals. The line $x=1$ appears in gold.
