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

Solving the first equation for $x$
$$
3x^2+\sin(2y)-\cos(y)-3=0
$$
$$
x=\pm\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}
$$
Substituting this into the second equation
$$
\left(\pm\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}\right)^3\mp3\sqrt{\frac{3+\cos(y)-\sin(2y)}{3}}-\sin(y)-\cos(2y)+3=0
$$
Simplifying radicals
$$
\pm\frac{1}{3^{3/2}}(3+\cos(y)-\sin(2y))^{3/2}\mp\frac{3}{\sqrt{3}}(3+\cos(y)-\sin(2y))^{1/2}-\sin(y)-\cos(2y)+3=0
$$
Expanding $\sin$'s, factoring the first two terms & ignoring $\pm,\mp$ signs to just take positive roots as this is already messy enough
$$
\frac{1}{\sqrt{3}}(3+\cos(y)-2\sin(y)\cos(y))^{1/2}\left(-2+\frac{1}{3}\cos(y)-\frac{2}{3}\sin(y)\cos(y)\right)-\sin(y)-\cos^2(y)+\sin^2(y)+3=0
$$
Substituting $a=\sin(y)$, $b=\cos(y)$
$$
\frac{1}{\sqrt{3}}(3+b-2ab)^{1/2}\left(\frac{1}{3}b-\frac{2}{3}ab-2\right)-a-b^2+a^2+3=0
$$
We arrive at this messy (but solvable) equation. I can't think of an efficient way to solve it but solutions for the variables $a$ & $b$ can be computed with Wolfram Alpha at the following link
$\quad$  $\quad$  $\quad$  $\quad$  $\quad$  $\quad$  $\quad$ Solutions $\quad$ (wait a minute for it to load)
If you substitute the original forms of $a$ & $b$ in terms of $y$ each solution gives an equation of the form $\cos(y)=f(\sin(y))$
Sorry i couldn't come up with anything neater nor more general but the important thing is that a solution can be found. Also, I may be being ignorant & not noticing an easier solution. If there are any apparent errors in my work, please let me know. Thanks for the interesting question.
